From 6cff327e88db3ca4e8db4939feb0a980cbd0e667 Mon Sep 17 00:00:00 2001 From: Maxim Dolgolyov Date: Sat, 16 May 2026 12:53:49 +0300 Subject: [PATCH] =?UTF-8?q?feat:=20exam9=20=E2=80=94=20=D0=AD=D0=BA=D0=B7?= =?UTF-8?q?=D0=B0=D0=BC=D0=B5=D0=BD=209=20=D0=BA=D0=BB=D0=B0=D1=81=D1=81?= =?UTF-8?q?=20=D0=BF=D0=BE=20=D0=BC=D0=B0=D1=82=D0=B5=D0=BC=D0=B0=D1=82?= =?UTF-8?q?=D0=B8=D0=BA=D0=B5=20(80=20=D0=B2=D0=B0=D1=80=D0=B8=D0=B0=D0=BD?= =?UTF-8?q?=D1=82=D0=BE=D0=B2)?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Новый отдельный модуль /exam9 в стиле LearnSpace: - 80 вариантов × 10 заданий = 800 задач с разбором (KaTeX + SVG) - Сайдбар: пункт «Экзамен 9 класс» (clipboard-check) - Feature flag: feature_exam9_enabled (мигр. 002) - Видим всем авторизованным; рендер на стороне клиента - Прогресс в localStorage: подсветка вариантов (done/partial) - Возобновление последнего варианта при возврате Структура: frontend/exam9.html — страница (LearnSpace layout) frontend/js/exam9/app.js — рендерер frontend/js/exam9/variants/ — 80 файлов с данными frontend/img/exam9/ — 22 PNG/JPG фигур заданий Картинки путей _tmp/ → /img/exam9/ переписаны автоматически. 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Экзамен 9 класс — Математика
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2025 · 80 вариантов · решения с разбором
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z9|4}{X+*u9k+k#NZX+>Ibh6pCT$x;uJuAq~{6U3gm9R-QIPAx`oeDz%;$KS5v( z5xC)lwh>-;GKK}zX-=~Nz~MZ>z=8${dJ(w^$kz%7N6fTlXRS@X;&A**i$n;a+)kuF z#HRRAq*CGeM|r}cnqXHF1s!?@wYF1cQdhC_ds3%AC#*8KDX z-20NmgXYizau0Zar()Ll5q$b|G*!z#V;NXaH?8* z!kqXkpSNhfm4=OlL-_!32V3d4!(D_0J$J~;nY<1_EJL;FWxHHwHU1k{;Ju-@J81ct zY2K-y3Z?^GvxUaaINyFdBe%T@97YmDZJ+izuxTo@;)xQ+IVyrwcF&Y&ipK>d!;_xI zzY1(LwJmZa+}tC$MV-}I{phKbH`k^rZKW|SdnzS;>i4{_oFji_b z;MKiCY4HGW!|+yrZ=6#YBr^D;3<~9ew^Mxx9U?|6>wtrwz7G@8@4^sceo@chfJxV$`hu0mInv0{Y!4+wlJf D7NLXz literal 0 HcmV?d00001 diff --git a/frontend/js/exam9/app.js b/frontend/js/exam9/app.js new file mode 100644 index 0000000..d04e6c0 --- /dev/null +++ b/frontend/js/exam9/app.js @@ -0,0 +1,176 @@ +'use strict'; +/* ────────────────────────────────────────────────────────────────── + Exam 9 — Math 2025 renderer + Variants loaded into window.VARIANTS by /js/exam9/variants/vNN.js + ────────────────────────────────────────────────────────────────── */ + +const STORAGE_KEY = 'exam9_progress_v1'; +let currentVariant = null; +let katexLoaded = false; + +/* ── KaTeX bootstrap ────────────────────────────────────────────── */ +function onKatexLoad() { + katexLoaded = true; + if (currentVariant !== null) runKatex(document.getElementById('ex-main')); +} + +function runKatex(el) { + if (!katexLoaded || !el) return; + try { + renderMathInElement(el, { + delimiters: [ + { left: '$$', right: '$$', display: true }, + { left: '$', right: '$', display: false }, + ], + throwOnError: false, + }); + } catch {} +} + +/* ── Progress in localStorage ───────────────────────────────────── */ +function loadProgress() { + try { return JSON.parse(localStorage.getItem(STORAGE_KEY) || '{}'); } + catch { return {}; } +} +function saveProgress(p) { + try { localStorage.setItem(STORAGE_KEY, JSON.stringify(p)); } catch {} +} +function markSolutionViewed(variantNum, taskIdx) { + const p = loadProgress(); + p[variantNum] = p[variantNum] || []; + if (!p[variantNum].includes(taskIdx)) { + p[variantNum].push(taskIdx); + saveProgress(p); + } +} + +/* ── Variant picker ─────────────────────────────────────────────── */ +function buildGrid() { + const grid = document.getElementById('variant-grid'); + const progress = loadProgress(); + grid.innerHTML = ''; + Object.keys(VARIANTS).sort((a, b) => Number(a) - Number(b)).forEach(n => { + const v = VARIANTS[n]; + const total = (v.tasks || []).length; + const viewed = (progress[n] || []).length; + let cls = ''; + if (viewed === total && total > 0) cls = ' done'; + else if (viewed > 0) cls = ' partial'; + const isActive = Number(n) === currentVariant ? ' active' : ''; + + const btn = document.createElement('button'); + btn.className = 'vg-btn' + cls + isActive; + btn.textContent = n; + btn.title = `${v.label}${viewed === total ? ' ✓' : viewed > 0 ? ` (${viewed}/${total})` : ''}`; + btn.onclick = () => { selectVariant(Number(n)); closePicker(); }; + grid.appendChild(btn); + }); +} + +function togglePicker() { + const overlay = document.getElementById('picker-overlay'); + const btn = document.getElementById('picker-btn'); + if (overlay.classList.contains('visible')) closePicker(); + else { + buildGrid(); + overlay.classList.add('visible'); + btn.classList.add('open'); + document.addEventListener('keydown', onEsc); + } +} + +function closePicker() { + document.getElementById('picker-overlay').classList.remove('visible'); + document.getElementById('picker-btn').classList.remove('open'); + document.removeEventListener('keydown', onEsc); +} + +function onOverlayClick(e) { + if (e.target === document.getElementById('picker-overlay')) closePicker(); +} +function onEsc(e) { if (e.key === 'Escape') closePicker(); } + +/* ── Task rendering ─────────────────────────────────────────────── */ +function buildOpts(opts) { + const isLong = opts.some(([, t]) => t.length > 40 && !t.startsWith('$')); + const cls = isLong ? 'opts-vertical' : 'opts'; + return `

` + + opts.map(([l, t]) => + `${l})${t}` + ).join('') + `
`; +} + +const SOL_ICON_CLOSED = ``; + +function renderVariant(num) { + const main = document.getElementById('ex-main'); + const v = VARIANTS[num]; + if (!v) { + main.innerHTML = '
Вариант не найден
'; + return; + } + + main.innerHTML = + `
${v.label}${v.tasks.length} заданий
` + + v.tasks.map((t, i) => ` +
+
+
${i + 1}
+
Задание ${i + 1}
+
+
+
${t.text}
+ ${t.figure ? `
${t.figure}
` : ''} + ${t.opts ? buildOpts(t.opts) : ''} +
+ ${t.sol ? `
+ +
${t.sol}
+
` : ''} +
` + ).join(''); + + runKatex(main); +} + +function toggleSol(btn, variantNum, taskIdx) { + const panel = btn.nextElementSibling; + const open = panel.classList.contains('visible'); + panel.classList.toggle('visible', !open); + btn.classList.toggle('open', !open); + btn.querySelector('span').textContent = open ? 'Показать решение' : 'Скрыть решение'; + if (!open) { + if (!panel.dataset.k) { runKatex(panel); panel.dataset.k = '1'; } + markSolutionViewed(variantNum, taskIdx); + } +} + +function selectVariant(num) { + currentVariant = num; + document.getElementById('picker-label').textContent = VARIANTS[num].label; + document.querySelectorAll('.vg-btn').forEach(b => { + b.classList.toggle('active', Number(b.textContent) === num); + }); + renderVariant(num); + // Persist last opened variant + try { localStorage.setItem('exam9_last_variant', String(num)); } catch {} + window.scrollTo({ top: 0, behavior: 'smooth' }); +} + +/* ── Boot ───────────────────────────────────────────────────────── */ +(function boot() { + const keys = Object.keys(VARIANTS); + if (!keys.length) { + document.getElementById('ex-main').innerHTML = '
Варианты не загружены
'; + return; + } + // Resume last opened variant or open first one + let initial = Number(keys[0]); + try { + const last = Number(localStorage.getItem('exam9_last_variant')); + if (last && VARIANTS[last]) initial = last; + } catch {} + selectVariant(initial); +})(); diff --git a/frontend/js/exam9/variants/v01.js b/frontend/js/exam9/variants/v01.js new file mode 100644 index 0000000..5060a93 --- /dev/null +++ b/frontend/js/exam9/variants/v01.js @@ -0,0 +1,203 @@ +VARIANTS[1] = { + label: "Вариант 1", + tasks: [ + { + text: `Какое из следующих чисел является натуральным:`, + opts: [ + ["а", "$-1$"], ["б", "$0$"], ["в", "$1{,}65$"], + ["г", "$36$"], ["д", "$\\dfrac{9}{50}$"], + ], + sol: `Натуральные числа — это 1, 2, 3, … (положительные целые). +
    +
  • а) $-1$ — отрицательное;
    • +
  • б) $0$ — нуль, не натуральное;
    • +
  • в) $1{,}65$ — дробное;
    • +
  • г) $36 = 6^2$ — натуральное ✓
    • +
  • д) $\\dfrac{9}{50}$ — правильная дробь.
    • +
+
Ответ: г) $36$
` + }, + { + text: `Результат упрощения выражения $5a^4 : a^{-15}$ имеет вид:`, + opts: [ + ["а", "$5a^{-11}$"], ["б", "$5a^{19}$"], ["в", "$\\dfrac{5}{a^{19}}$"], + ["г", "$5a^{11}$"], ["д", "$\\dfrac{a^{11}}{5}$"], + ], + sol: `При делении степеней с одним основанием показатели вычитаются: +$$5a^4 : a^{-15} = 5\\cdot a^{4-(-15)} = 5\\cdot a^{4+15} = 5a^{19}$$ +
Ответ: б) $5a^{19}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "вертикальные углы равны между собой;"], + ["б", "если в треугольнике два угла равны, то он равнобедренный;"], + ["в", "противоположные стороны параллелограмма равны;"], + ["г", "диагонали любого ромба равны между собой?"], + ], + sol: `
    +
  • а) Вертикальные углы равны — верно
    • +
  • б) Два равных угла ⟹ равнобедренный — верно
    • +
  • в) Противоположные стороны параллелограмма равны — верно
    • +
  • г) Диагонали любого ромба равны — НЕВЕРНО
    • +
+В ромбе диагонали взаимно перпендикулярны и делятся пополам, но равны по длине лишь у квадрата — частного случая ромба. +
Ответ: г)
` + }, + { + text: `Определите наименьшее целое решение совокупности неравенств + $$\\left[\\begin{array}{l} x^2 - 3x \\leq 0, \\\\[4pt] x > -2{,}5. \\end{array}\\right.$$`, + sol: `Совокупность «$[\\,$» означает объединение: решение удовлетворяет хотя бы одному из неравенств. +
Неравенство 1: $x^2-3x\\leq 0\\Rightarrow x(x-3)\\leq 0\\Rightarrow 0\\leq x\\leq 3$ +
Неравенство 2: $x>-2{,}5$ +
Объединение: $(0\\leq x\\leq 3)\\cup(x>-2{,}5) = x>-2{,}5$ +−2−10123−2,5Наименьшее целое число, большее $-2{,}5$: это $-2$. +
Ответ: $-2$
` + }, + { + text: `Сократите дробь $\\dfrac{x^2 - 12x + 35}{x^2 - 10x + 25}$ + и найдите значение полученной дроби при $x = 3$.`, + sol: `Теорема Виета (обратная): если $x_1+x_2=-b/a$ и $x_1\\cdot x_2=c/a$, то $ax^2+bx+c=a(x-x_1)(x-x_2)$. +
Формула квадрата суммы: $(x\\pm a)^2 = x^2\\pm 2ax+a^2$. +

+Шаг 1. Разложим числитель на множители. Ищем такие $x_1$ и $x_2$, что $x_1+x_2=12$ и $x_1\\cdot x_2=35$. +
Подходят $x_1=5$ и $x_2=7$, поэтому: +$$x^2-12x+35 = (x-5)(x-7)$$ +Шаг 2. Разложим знаменатель. Замечаем, что это полный квадрат: +$$x^2-10x+25 = x^2-2\\cdot 5\\cdot x+5^2 = (x-5)^2$$ +Шаг 3. Сокращаем общий множитель $(x-5)$ (так как $x\\neq 5$, иначе знаменатель обращается в нуль): +$$\\dfrac{(x-5)(x-7)}{(x-5)^2} = \\dfrac{x-7}{x-5}$$ +Шаг 4. Подставляем $x=3$ в полученную дробь: +$$\\dfrac{3-7}{3-5} = \\dfrac{-4}{-2} = 2$$ +
Ответ: $2$
` + }, + { + text: `Диагонали параллелограмма $ABCD$ пересекаются в точке $O$, $BC = 10$ см. + Высота, проведённая из вершины $C$ к стороне $AD$, равна $6$ см. + Найдите площадь треугольника $AOB$.`, + sol: `Свойства параллелограмма: противоположные стороны равны; диагонали точкой пересечения делятся пополам и разбивают параллелограмм на 4 равновеликих треугольника. +
Формула площади параллелограмма: $S = a\\cdot h$, где $a$ — сторона, $h$ — высота к ней. +

+Шаг 1. Так как $BC$ и $AD$ — противоположные стороны параллелограмма, то $AD = BC = 10$ см. +
Шаг 2. Высота из $C$ к $AD$ равна $6$ см, поскольку $BC\\parallel AD$, значит высота от стороны $BC$ к $AD$ — это расстояние между параллельными прямыми. +
Шаг 3. Находим площадь всего параллелограмма: +$$S_{ABCD} = AD\\cdot h = 10\\cdot 6 = 60\\text{ см}^2$$ +Шаг 4. Точка $O$ — пересечение диагоналей. Диагонали разбивают параллелограмм на четыре треугольника, у которых равные площади (каждая равна $S/4$): +ABCDOS/4 +Шаг 5. Поэтому: +$$S_{AOB} = \\dfrac{S_{ABCD}}{4} = \\dfrac{60}{4} = 15\\text{ см}^2$$ +
Ответ: $15$ см²
` + }, + { + text: `Отчисления в бюджет по фиксированной ставке с доходов физических лиц + в Беларуси составляют $13\\%$ от заработной платы. После удержания налога + на доходы сотрудник предприятия получил $1305$ р. + Сколько рублей составляет заработная плата сотрудника без вычета налога?`, + sol: `Метод уравнения для задачи на проценты: неизвестную величину обозначаем переменной, выражаем её через известные данные и составляем уравнение. +
Свойство: если из величины удерживают $p\\%$, то остаётся $(100-p)\\%$. +

+Шаг 1. Обозначим за $x$ заработную плату до удержания налога — это и есть то, что мы ищем. +
Шаг 2. По условию налог составляет $13\\%$ от $x$. Значит, на руки сотрудник получает оставшиеся $100\\%-13\\%=87\\%$ от $x$: +$$0{,}87\\cdot x = 1305$$ +Шаг 3. Решаем уравнение — делим обе части на $0{,}87$: +$$x = \\dfrac{1305}{0{,}87} = \\dfrac{1305\\cdot 100}{87} = \\dfrac{130500}{87}$$ +Шаг 4. Выполняем деление: $130500:87 = 1500$. +$$x = 1500\\text{ р.}$$ +Проверка: $13\\%$ от $1500$ — это $0{,}13\\cdot 1500=195$ р.; на руки $1500-195=1305$ р. ✓ +
Ответ: $1500$ р.
` + }, + { + text: `Найдите значение выражения $x_1 \\cdot x_2 + y_1 \\cdot y_2$, + где $(x_1;\\, y_1)$, $(x_2;\\, y_2)$ — решения системы уравнений + $$\\begin{cases} x^2 - y = 21, \\\\[4pt] x + y = 9. \\end{cases}$$`, + sol: `Метод подстановки для системы уравнений: выражаем одну переменную через другую и подставляем. +
Теорема Виета (обратная): для $x^2+px+q=0$ корни $x_1,x_2$ удовлетворяют $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. Из второго уравнения системы выразим $y$ через $x$: +$$y = 9 - x$$ +Шаг 2. Подставим это выражение в первое уравнение: +$$x^2 - (9-x) = 21$$ +$$x^2 + x - 30 = 0$$ +Шаг 3. Решаем квадратное уравнение. По теореме Виета: $x_1+x_2=-1$, $x_1\\cdot x_2=-30$. Подходят $-6$ и $5$: +$$(x+6)(x-5)=0 \\implies x_1=-6,\\; x_2=5$$ +Шаг 4. Для каждого корня находим соответствующий $y$ по формуле $y=9-x$: +
$x_1=-6$:$y_1=9-(-6)=15$
$x_2=5$:$y_2=9-5=4$
+Шаг 5. Вычисляем нужное выражение: +$$x_1 x_2 + y_1 y_2 = (-6)\\cdot 5 + 15\\cdot 4 = -30+60 = 30$$ +
Ответ: $30$
` + }, + { + text: `Функция $y = f(x)$ нечётная и для $x > 0$ задаётся формулой + $f(x) = -x^2 - \\dfrac{1}{x}$. + Найдите значение выражения $f(-2) - f\\!\\left(-\\dfrac{1}{2}\\right)$.`, + sol: `Свойство нечётной функции: $f(-x)=-f(x)$ для всех $x$ из области определения. +
Идея решения: формула $f(x)=-x^2-\\dfrac{1}{x}$ дана только при $x\\gt 0$. Чтобы найти значения функции в отрицательных точках, используем свойство нечётности: $f(-a)=-f(a)$. +

+Шаг 1. Вычисляем $f(2)$ по данной формуле (так как $2\\gt 0$): +$$f(2) = -2^2 - \\dfrac{1}{2} = -4 - \\dfrac{1}{2} = -\\dfrac{9}{2}$$ +Шаг 2. Так как функция нечётная, то $f(-2) = -f(2)$: +$$f(-2) = -\\left(-\\dfrac{9}{2}\\right) = \\dfrac{9}{2}$$ +Шаг 3. Вычисляем $f\\!\\left(\\dfrac{1}{2}\\right)$ по той же формуле (так как $\\dfrac{1}{2}\\gt 0$): +$$f\\!\\left(\\dfrac{1}{2}\\right) = -\\left(\\dfrac{1}{2}\\right)^2 - \\dfrac{1}{1/2} = -\\dfrac{1}{4} - 2 = -\\dfrac{9}{4}$$ +Шаг 4. По свойству нечётности: +$$f\\!\\left(-\\dfrac{1}{2}\\right) = -f\\!\\left(\\dfrac{1}{2}\\right) = -\\left(-\\dfrac{9}{4}\\right) = \\dfrac{9}{4}$$ +Шаг 5. Находим требуемую разность: +$$f(-2) - f\\!\\left(-\\dfrac{1}{2}\\right) = \\dfrac{9}{2} - \\dfrac{9}{4} = \\dfrac{18}{4} - \\dfrac{9}{4} = \\dfrac{9}{4}$$ +
Ответ: $\\dfrac{9}{4}$
` + }, + { + text: `В прямоугольном треугольнике точка касания вписанной окружности + делит гипотенузу на отрезки длиной $5$ см и $3$ см. + Найдите площадь треугольника.`, + sol: `Обозначения. Пусть прямой угол в точке $C$, катеты $CA=b$, $CB=a$, гипотенуза $AB=c$. Вписанная окружность имеет радиус $r$ и касается гипотенузы в точке $P$: $AP=5$, $PB=3$. + + + + + + + + + + + + A + B + C + P + r + r + 5 + 3 + 5 + r + r + +Шаг 1. Свойство касательных. +
Из каждой вершины отрезки до двух точек касания равны. Поэтому: +
    +
  • Из $A$: касательная к гипотенузе $AP=5$, касательная к катету $CA$ тоже $=5$.
    • +
  • Из $B$: касательная к гипотенузе $BP=3$, касательная к катету $CB$ тоже $=3$.
    • +
  • Из $C$ (прямой угол): обе касательные равны $r$.
    • +
+Значит: +$$AB = 5+3 = 8\\text{ см}, \\quad CA = 5+r, \\quad CB = 3+r$$ +Шаг 2. Теорема Пифагора. +$$CA^2 + CB^2 = AB^2$$ +$$(5+r)^2 + (3+r)^2 = 8^2$$ +Раскрываем скобки: +$$(25+10r+r^2)+(9+6r+r^2)=64$$ +$$2r^2+16r+34=64$$ +$$2r^2+16r-30=0$$ +$$r^2+8r-15=0 \\quad (*)$$ +Шаг 3. Площадь без нахождения $r$. +
Площадь прямоугольного треугольника: +$$S=\\dfrac{1}{2}\\cdot CA\\cdot CB = \\dfrac{1}{2}(5+r)(3+r)$$ +Раскрываем скобки: +$$S=\\dfrac{1}{2}\\bigl(15+8r+r^2\\bigr)$$ +Из уравнения $(*)$: $r^2+8r=15$. Подставляем: +$$S=\\dfrac{1}{2}\\bigl(15+15\\bigr)=\\dfrac{1}{2}\\cdot30=15\\text{ см}^2$$ +
Ответ: $15$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v02.js b/frontend/js/exam9/variants/v02.js new file mode 100644 index 0000000..3edce35 --- /dev/null +++ b/frontend/js/exam9/variants/v02.js @@ -0,0 +1,197 @@ +VARIANTS[2] = { + label: "Вариант 2", + tasks: [ + { + text: `Какое из следующих чисел является натуральным:`, + opts: [ + ["а", "$-6$"], ["б", "$0$"], ["в", "$2{,}5$"], + ["г", "$\\dfrac{7}{30}$"], ["д", "$143$"], + ], + sol: `Натуральные числа — это 1, 2, 3, … (положительные целые). +
    +
  • а) $-6$ — отрицательное;
    • +
  • б) $0$ — нуль, не натуральное;
    • +
  • в) $2{,}5$ — дробное;
    • +
  • г) $\\dfrac{7}{30}$ — правильная дробь;
    • +
  • д) $143$ — натуральное ✓
    • +
+
Ответ: д) $143$
` + }, + { + text: `Результат упрощения выражения $4a^6 : a^{-12}$ имеет вид:`, + opts: [ + ["а", "$4a^{-6}$"], ["б", "$4a^6$"], ["в", "$\\dfrac{4}{a^{18}}$"], + ["г", "$4a^{18}$"], ["д", "$\\dfrac{a^6}{4}$"], + ], + sol: `При делении степеней с одним основанием показатели вычитаются: +$$4a^6 : a^{-12} = 4\\cdot a^{6-(-12)} = 4\\cdot a^{6+12} = 4a^{18}$$ +
Ответ: г) $4a^{18}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "в равнобедренном треугольнике углы при основании равны;"], + ["б", "сумма смежных углов равна $180^{\\circ}$;"], + ["в", "у любого параллелограмма все стороны равны;"], + ["г", "диагонали любого ромба перпендикулярны?"], + ], + sol: `
    +
  • а) Углы при основании равнобедренного треугольника равны — верно
    • +
  • б) Сумма смежных углов равна $180°$ — верно
    • +
  • в) У любого параллелограмма все стороны равны — НЕВЕРНО
    • +
  • г) Диагонали любого ромба перпендикулярны — верно
    • +
+Все стороны равны лишь у ромба — частного случая параллелограмма; в общем параллелограмме противоположные стороны равны, но смежные стороны могут различаться. +
Ответ: в)
` + }, + { + text: `Определите наименьшее целое решение совокупности неравенств + $$\\left[\\begin{array}{l} x^2 - 4x \\leq 0, \\\\[4pt] x > -1{,}5. \\end{array}\\right.$$`, + sol: `Совокупность «$[\\,$» означает объединение: решение удовлетворяет хотя бы одному из неравенств. +
Неравенство 1: $x^2-4x\\leq 0\\Rightarrow x(x-4)\\leq 0\\Rightarrow 0\\leq x\\leq 4$ +
Неравенство 2: $x>-1{,}5$ +
Объединение: $(0\\leq x\\leq 4)\\cup(x>-1{,}5) = x>-1{,}5$ +−1012−1,5Наименьшее целое число, большее $-1{,}5$: это $-1$. +
Ответ: $-1$
` + }, + { + text: `Сократите дробь $\\dfrac{x^2 + x - 12}{x^2 + 8x + 16}$ + и найдите значение полученной дроби при $x = -3$.`, + sol: `Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +
Формула квадрата суммы: $(x+a)^2 = x^2+2ax+a^2$. +

+Шаг 1. Разложим числитель на множители. Ищем числа с суммой $-1$ и произведением $-12$. Подходят $-3$ и $4$: +$$x^2+x-12 = (x+4)(x-3)$$ +Шаг 2. Разложим знаменатель. Это полный квадрат: +$$x^2+8x+16 = x^2+2\\cdot 4\\cdot x + 4^2 = (x+4)^2$$ +Шаг 3. Сокращаем общий множитель $(x+4)$ (так как $x\\neq -4$, иначе знаменатель равен нулю): +$$\\dfrac{(x+4)(x-3)}{(x+4)^2} = \\dfrac{x-3}{x+4}$$ +Шаг 4. Подставляем $x=-3$: +$$\\dfrac{-3-3}{-3+4} = \\dfrac{-6}{1} = -6$$ +
Ответ: $-6$
` + }, + { + text: `Диагонали параллелограмма $ABCD$ пересекаются в точке $O$, $AD = 8$ см. + Высота, проведённая из вершины $A$ к стороне $BC$, равна $4$ см. + Найдите площадь треугольника $AOD$.`, + sol: `Свойства параллелограмма: противоположные стороны равны; диагонали точкой пересечения делятся пополам и разбивают параллелограмм на 4 равновеликих треугольника. +
Формула площади параллелограмма: $S = a\\cdot h$, где $a$ — сторона, $h$ — высота к ней. +

+Шаг 1. Так как $AD$ и $BC$ — противоположные стороны параллелограмма, то $BC = AD = 8$ см. +
Шаг 2. Высота из вершины $A$ к стороне $BC$ длиной $4$ см — это расстояние между параллельными прямыми $AD$ и $BC$. +
Шаг 3. Находим площадь всего параллелограмма: +$$S_{ABCD} = AD\\cdot h = 8\\cdot 4 = 32\\text{ см}^2$$ +Шаг 4. Точка $O$ — пересечение диагоналей. Диагонали делят параллелограмм на четыре равновеликих треугольника: +ABCDOS/4 +Шаг 5. Значит: +$$S_{AOD} = \\dfrac{S_{ABCD}}{4} = \\dfrac{32}{4} = 8\\text{ см}^2$$ +
Ответ: $8$ см²
` + }, + { + text: `Отчисления в бюджет по фиксированной ставке с доходов физических лиц + в Беларуси составляют $13\\%$ от заработной платы. После удержания налога + на доходы сотрудник предприятия получил $1131$ р. + Сколько рублей составляет заработная плата сотрудника без вычета налога?`, + sol: `Метод уравнения для задачи на проценты: неизвестную величину обозначаем переменной и выражаем условие задачи как уравнение. +
Свойство: если из величины удерживают $p\\%$, то остаётся $(100-p)\\%$. +

+Шаг 1. Обозначим за $x$ заработную плату сотрудника до удержания налога. +
Шаг 2. По условию налог $= 13\\%$ от $x$. Значит, на руки сотрудник получает оставшиеся $100\\%-13\\%=87\\%$ от $x$: +$$0{,}87\\cdot x = 1131$$ +Шаг 3. Делим обе части уравнения на $0{,}87$: +$$x = \\dfrac{1131}{0{,}87} = \\dfrac{1131\\cdot 100}{87} = \\dfrac{113100}{87}$$ +Шаг 4. Выполняем деление: $113100:87 = 1300$. +$$x = 1300\\text{ р.}$$ +Проверка: $13\\%$ от $1300$ — это $0{,}13\\cdot 1300=169$ р.; на руки $1300-169=1131$ р. ✓ +
Ответ: $1300$ р.
` + }, + { + text: `Найдите значение выражения $x_1 \\cdot x_2 + y_1 \\cdot y_2$, + где $(x_1;\\, y_1)$, $(x_2;\\, y_2)$ — решения системы уравнений + $$\\begin{cases} x^2 - y = 16, \\\\[4pt] x + y = 4. \\end{cases}$$`, + sol: `Метод подстановки для системы уравнений: выражаем одну переменную через другую и подставляем. +
Теорема Виета (обратная): для $x^2+px+q=0$ корни $x_1,x_2$ удовлетворяют $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. Из второго уравнения выразим $y$ через $x$: +$$y = 4 - x$$ +Шаг 2. Подставим в первое уравнение: +$$x^2 - (4-x) = 16$$ +$$x^2 + x - 20 = 0$$ +Шаг 3. По теореме Виета ищем корни: $x_1+x_2=-1$, $x_1\\cdot x_2=-20$. Подходят $-5$ и $4$: +$$(x+5)(x-4)=0 \\implies x_1=-5,\\; x_2=4$$ +Шаг 4. По формуле $y = 4-x$ находим $y$ для каждого корня: +
$x_1=-5$:$y_1=4-(-5)=9$
$x_2=4$:$y_2=4-4=0$
+Шаг 5. Вычисляем требуемое выражение: +$$x_1 x_2 + y_1 y_2 = (-5)\\cdot 4 + 9\\cdot 0 = -20+0 = -20$$ +
Ответ: $-20$
` + }, + { + text: `Функция $y = f(x)$ нечётная и для $x > 0$ задаётся формулой + $f(x) = \\dfrac{1}{x} - x^2$. + Найдите значение выражения $f\\!\\left(-\\dfrac{1}{3}\\right) - f(-3)$.`, + sol: `Свойство нечётной функции: $f(-x)=-f(x)$ для всех $x$ из области определения. +
Идея решения: формула задана только при $x\\gt 0$. Чтобы найти значения функции в отрицательных точках, применяем свойство нечётности. +

+Шаг 1. Вычислим $f\\!\\left(\\dfrac{1}{3}\\right)$ по данной формуле (так как $\\dfrac{1}{3}\\gt 0$): +$$f\\!\\left(\\dfrac{1}{3}\\right) = \\dfrac{1}{1/3} - \\left(\\dfrac{1}{3}\\right)^2 = 3 - \\dfrac{1}{9} = \\dfrac{27-1}{9} = \\dfrac{26}{9}$$ +Шаг 2. По свойству нечётности: +$$f\\!\\left(-\\dfrac{1}{3}\\right) = -f\\!\\left(\\dfrac{1}{3}\\right) = -\\dfrac{26}{9}$$ +Шаг 3. Вычислим $f(3)$ по той же формуле: +$$f(3) = \\dfrac{1}{3} - 3^2 = \\dfrac{1}{3} - 9 = \\dfrac{1-27}{3} = -\\dfrac{26}{3}$$ +Шаг 4. По свойству нечётности: +$$f(-3) = -f(3) = -\\left(-\\dfrac{26}{3}\\right) = \\dfrac{26}{3}$$ +Шаг 5. Находим требуемую разность. Приведём дроби к общему знаменателю $9$: +$$f\\!\\left(-\\dfrac{1}{3}\\right) - f(-3) = -\\dfrac{26}{9} - \\dfrac{26}{3} = -\\dfrac{26}{9} - \\dfrac{78}{9} = -\\dfrac{104}{9}$$ +
Ответ: $-\\dfrac{104}{9}$
` + }, + { + text: `В прямоугольном треугольнике точка касания вписанной окружности + делит гипотенузу на отрезки длиной $4$ см и $3$ см. + Найдите площадь треугольника.`, + sol: `Обозначения. Пусть прямой угол в точке $C$, катеты $CA=b$, $CB=a$, гипотенуза $AB=c$. Вписанная окружность касается гипотенузы в точке $P$: $AP=4$, $PB=3$. + + + + + + + + + + + + A + B + C + P + r + r + 4 + 3 + 4 + r + r + +Шаг 1. Свойство касательных. +
Из каждой вершины отрезки до двух точек касания равны: +
    +
  • Из $A$: касательная к гипотенузе $AP=4$, касательная к катету $CA$ тоже $=4$.
    • +
  • Из $B$: касательная к гипотенузе $BP=3$, касательная к катету $CB$ тоже $=3$.
    • +
  • Из $C$ (прямой угол): обе касательные равны $r$.
    • +
+Значит: +$$AB = 4+3 = 7\\text{ см}, \\quad CA = 4+r, \\quad CB = 3+r$$ +Шаг 2. Теорема Пифагора. +$$(4+r)^2 + (3+r)^2 = 7^2$$ +$$16+8r+r^2+9+6r+r^2=49$$ +$$2r^2+14r+25=49$$ +$$2r^2+14r-24=0$$ +$$r^2+7r-12=0 \\quad (*)$$ +Шаг 3. Площадь без нахождения $r$. +$$S=\\dfrac{1}{2}(4+r)(3+r)=\\dfrac{1}{2}\\bigl(12+7r+r^2\\bigr)$$ +Из уравнения $(*)$: $r^2+7r=12$. Подставляем: +$$S=\\dfrac{1}{2}\\bigl(12+12\\bigr)=\\dfrac{1}{2}\\cdot24=12\\text{ см}^2$$ +
Ответ: $12$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v03.js b/frontend/js/exam9/variants/v03.js new file mode 100644 index 0000000..049c2cb --- /dev/null +++ b/frontend/js/exam9/variants/v03.js @@ -0,0 +1,231 @@ +VARIANTS[3] = { + label: "Вариант 3", + tasks: [ + { + text: `Определите, какое из следующих равенств верно:`, + opts: [ + ["а", "$a^2 \\cdot a^8 = a^{10}$"], ["б", "$a^2 \\cdot a^8 = 10a$"], + ["в", "$a^2 \\cdot a^8 = a^{16}$"], ["г", "$a^2 \\cdot a^8 = a^6$"], + ["д", "$a^2 \\cdot a^8 = a^{64}$"], + ], + sol: `При умножении степеней с одним основанием показатели складываются: +$$a^2 \\cdot a^8 = a^{2+8} = a^{10}$$ +
Ответ: а) $a^{10}$
` + }, + { + text: `Значение выражения $17\\dfrac{11}{23} - 5\\dfrac{14}{23}$ равно:`, + opts: [ + ["а", "$12\\dfrac{3}{23}$"], ["б", "$11\\dfrac{20}{23}$"], + ["в", "$12\\dfrac{20}{23}$"], ["г", "$11\\dfrac{19}{23}$"], + ["д", "$13\\dfrac{1}{23}$"], + ], + sol: `$$17\\tfrac{11}{23} - 5\\tfrac{14}{23} = (17-5) + \\left(\\tfrac{11}{23}-\\tfrac{14}{23}\\right) = 12 - \\tfrac{3}{23} = 11\\tfrac{23}{23} - \\tfrac{3}{23} = 11\\tfrac{20}{23}$$ +
Ответ: б) $11\\dfrac{20}{23}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "на плоскости через точку можно провести только одну прямую, перпендикулярную данной;"], + ["б", "в равнобедренном треугольнике углы при основании равны;"], + ["в", "у прямоугольника диагонали равны между собой;"], + ["г", "сумма всех углов квадрата равна $180^{\\circ}$?"], + ], + sol: `
    +
  • а) Единственная перпендикуляр из точки к прямой — верно
    • +
  • б) Углы при основании равнобедренного треугольника равны — верно
    • +
  • в) Диагонали прямоугольника равны — верно
    • +
  • г) Сумма углов квадрата = $180°$ — НЕВЕРНО
    • +
+Квадрат — четырёхугольник, сумма его углов $= 4\\times 90^\\circ = 360^\\circ \\neq 180^\\circ$. +
Ответ: г)
` + }, + { + text: `Определите количество целых решений системы неравенств + $$\\begin{cases} x \\leq 7, \\\\[4pt] 3 - x < 0. \\end{cases}$$`, + sol: `Из второго неравенства: $3-x < 0 \\Rightarrow x > 3$. +
Система: $3 < x \\leq 7$ +
Целые числа: $4,\\ 5,\\ 6,\\ 7$ — ровно 4 числа. +
Ответ: 4
` + }, + { + text: `В окружность с радиусом $10$ см вписан треугольник, одна из сторон которого + является диаметром, а другая — равна $16$ см. Найдите площадь этого треугольника.`, + sol: `Теорема Фалеса (о вписанном угле): вписанный угол, опирающийся на диаметр, прямой ($90°$). +
Теорема Пифагора: в прямоугольном треугольнике $c^2 = a^2 + b^2$, где $c$ — гипотенуза. +
Формула площади прямоугольного треугольника: $S = \\dfrac{1}{2}\\cdot a\\cdot b$ (полупроизведение катетов). +

+Шаг 1. Одна сторона треугольника — диаметр окружности, значит: +$$d = 2R = 2\\cdot 10 = 20\\text{ см}$$ +Шаг 2. По теореме Фалеса вписанный угол, опирающийся на диаметр, равен $90°$. Значит, треугольник прямоугольный, а его гипотенуза $= 20$ см. + + + + +A +B +C +20 см +16 см + +Шаг 3. По условию один катет равен $16$ см. По теореме Пифагора находим второй катет: +$$b = \\sqrt{20^2-16^2} = \\sqrt{400-256} = \\sqrt{144} = 12\\text{ см}$$ +Шаг 4. Площадь прямоугольного треугольника — полупроизведение катетов: +$$S = \\tfrac{1}{2}\\cdot 16\\cdot 12 = 96\\text{ см}^2$$ +
Ответ: $96$ см²
` + }, + { + text: `Найдите значение выражения $\\dfrac{a - 7}{a - 2\\sqrt{7a} + 7}$ при $a = 28$.`, + sol: `Формула разности квадратов: $a^2-b^2 = (a-b)(a+b)$. +
Формула квадрата разности: $(a-b)^2 = a^2-2ab+b^2$. +
Идея: при работе с радикалами полезно представить $a$ как $(\\sqrt{a})^2$, чтобы увидеть структуру формул сокращённого умножения. +

+Шаг 1. Разложим знаменатель. Представим $a = (\\sqrt{a})^2$ и $7=(\\sqrt{7})^2$. Тогда: +$$a - 2\\sqrt{7a} + 7 = (\\sqrt{a})^2 - 2\\sqrt{a}\\cdot\\sqrt{7} + (\\sqrt{7})^2 = (\\sqrt{a}-\\sqrt{7})^2$$ +(это квадрат разности). +
Шаг 2. Разложим числитель по формуле разности квадратов: +$$a - 7 = (\\sqrt{a})^2 - (\\sqrt{7})^2 = (\\sqrt{a}-\\sqrt{7})(\\sqrt{a}+\\sqrt{7})$$ +Шаг 3. Сокращаем общий множитель $(\\sqrt{a}-\\sqrt{7})$: +$$\\dfrac{(\\sqrt{a}-\\sqrt{7})(\\sqrt{a}+\\sqrt{7})}{(\\sqrt{a}-\\sqrt{7})^2} = \\dfrac{\\sqrt{a}+\\sqrt{7}}{\\sqrt{a}-\\sqrt{7}}$$ +Шаг 4. Подставляем $a=28$. Заметим, что $\\sqrt{28} = \\sqrt{4\\cdot 7} = 2\\sqrt{7}$: +$$\\dfrac{\\sqrt{28}+\\sqrt{7}}{\\sqrt{28}-\\sqrt{7}} = \\dfrac{2\\sqrt{7}+\\sqrt{7}}{2\\sqrt{7}-\\sqrt{7}} = \\dfrac{3\\sqrt{7}}{\\sqrt{7}} = 3$$ +
Ответ: $3$
` + }, + { + text: `Функция задана формулой $f(x) = \\dfrac{5}{x}$. + Расположите в порядке возрастания $f(-3{,}5)$, $f(-10{,}3)$, $f(-\\sqrt{5})$. + Ответ обоснуйте.`, + sol: `Свойство гиперболы $y=\\dfrac{k}{x}$ при $k\\gt 0$: функция убывает на каждом из промежутков $(-\\infty;0)$ и $(0;+\\infty)$. +
Правило сравнения значений убывающей функции: чем больше аргумент, тем меньше значение функции (на промежутке убывания). +

+Шаг 1. Определим характер функции. У функции $f(x)=\\dfrac{5}{x}$ коэффициент $k=5\\gt 0$, поэтому на промежутке $(-\\infty;0)$ функция убывает. +
Шаг 2. Все три аргумента отрицательны, значит, лежат на промежутке убывания. Сравним их между собой. +
Оценим $\\sqrt{5}$: так как $2^2=4\\lt 5\\lt 9=3^2$, имеем $2\\lt\\sqrt{5}\\lt 3$, точнее $\\sqrt{5}\\approx 2{,}24$, поэтому $-\\sqrt{5}\\approx -2{,}24$. +
Расставим аргументы по возрастанию (от меньшего к большему на числовой прямой): +$$-10{,}3 \\lt -3{,}5 \\lt -\\sqrt{5}$$ +Шаг 3. Так как функция $f$ убывает, при увеличении аргумента значение $f$ уменьшается. Значит, неравенства между значениями функции имеют противоположный смысл: +$$f(-10{,}3) \\gt f(-3{,}5) \\gt f(-\\sqrt{5})$$ +Шаг 4. Перепишем «по возрастанию» (от меньшего к большему): +$$f(-\\sqrt{5}) \\lt f(-3{,}5) \\lt f(-10{,}3)$$ +Проверка вычислением: $f(-\\sqrt{5})=-\\dfrac{5}{\\sqrt{5}}=-\\sqrt{5}\\approx -2{,}24$; $f(-3{,}5)\\approx -1{,}43$; $f(-10{,}3)\\approx -0{,}49$. Действительно, $-2{,}24\\lt -1{,}43\\lt -0{,}49$ ✓. +
Ответ (по возрастанию): $f(-\\sqrt{5}) \\lt f(-3{,}5) \\lt f(-10{,}3)$
` + }, + { + text: `Найдите сумму всех двузначных натуральных чисел, + которые при делении на $13$ дают в остатке $7$.`, + sol: `Теорема о делении с остатком: если число $n$ при делении на $d$ даёт остаток $r$, то $n = d\\cdot k + r$, где $k$ — натуральное число или ноль. +
Формула суммы арифметической прогрессии: $S_n = \\dfrac{(a_1+a_n)\\cdot n}{2}$. +

+Шаг 1. Запишем общий вид искомых чисел. По условию число делится на $13$ с остатком $7$, поэтому имеет вид: +$$n = 13k + 7,\\quad k = 0, 1, 2, \\ldots$$ +Шаг 2. Найдём, какие значения $k$ дают двузначные числа. Должно выполняться: $10\\leq 13k+7\\leq 99$. +
Левое неравенство: $13k\\geq 3 \\Rightarrow k\\geq 1$ (так как $k$ — целое). +
Правое неравенство: $13k\\leq 92 \\Rightarrow k\\leq 7$ (так как $13\\cdot 7=91\\leq 92$, а $13\\cdot 8=104\\gt 92$). +
Итак, $k = 1, 2, 3, 4, 5, 6, 7$ — всего $7$ значений. +
Шаг 3. Выпишем все двузначные числа, удовлетворяющие условию: +$$20,\\ 33,\\ 46,\\ 59,\\ 72,\\ 85,\\ 98$$ +Это арифметическая прогрессия с первым членом $a_1=20$, разностью $d=13$ и числом членов $n=7$. Последний член $a_7=98$. +
Шаг 4. По формуле суммы арифметической прогрессии: +$$S = \\dfrac{(a_1+a_n)\\cdot n}{2} = \\dfrac{(20+98)\\cdot 7}{2} = \\dfrac{118\\cdot 7}{2} = 59\\cdot 7 = 413$$ +
Ответ: $413$
` + }, + { + text: `Внутри угла $A$, равного $60^{\\circ}$, взята точка $M$. + Расстояния от точки $M$ до сторон угла равны $4$ см и $8$ см. + Найдите расстояние от точки $M$ до вершины угла $A$.`, + sol: `Опустим перпендикуляры $MH_1=4$ и $MH_2=8$ на стороны угла. + + + + + + d=AM + + + + + + + + 60° + + A + + + M + + + + + 4 + H₁ + + + + + + + + + + + 8 + H₂ + + + + 4√7 + + + + + + + + 120° + + + O + +Шаг 1 — угол при M.
+Четырёхугольник $AH_1MH_2$ имеет три известных угла. Так как сумма углов любого четырёхугольника равна $360°$: +$$\\angle H_1MH_2 = 360° - \\underbrace{90°}_{\\angle H_1} - \\underbrace{60°}_{\\angle A} - \\underbrace{90°}_{\\angle H_2} = \\boldsymbol{120°}$$ +Шаг 2 — длина отрезка H₁H₂.
+Теорема косинусов в $\\triangle H_1MH_2$: +$$H_1H_2^2 = 4^2 + 8^2 - 2\\cdot4\\cdot8\\cdot\\cos120° = 16+64-64\\cdot\\left(-\\tfrac{1}{2}\\right) = 112$$ +$$H_1H_2 = 4\\sqrt{7}$$ +Шаг 3 — четыре точки на одной окружности.
+Так как $\\angle AH_1M = \\angle AH_2M = 90°$, по обратной теореме Фалеса: точки $H_1$ и $H_2$ лежат на окружности с диаметром $AM$.
+Итого все четыре точки $A, H_1, H_2, M$ вписаны в одну окружность, диаметр которой равен $AM$. +

+Шаг 4 — теорема синусов.
+В описанной окружности (диаметр $= AM$) вписанный угол $\\angle H_1AH_2 = 60°$ опирается на хорду $H_1H_2$. +По теореме синусов: +$$\\dfrac{H_1H_2}{\\sin\\angle H_1AH_2} = AM \\implies \\dfrac{4\\sqrt{7}}{\\sin 60°} = AM$$ +$$AM = \\dfrac{4\\sqrt{7}}{\\dfrac{\\sqrt{3}}{2}} = \\dfrac{8\\sqrt{7}}{\\sqrt{3}} = \\dfrac{8\\sqrt{21}}{3} \\approx 12{,}2\\text{ см}$$ +
Ответ: $\\dfrac{8\\sqrt{21}}{3}$ см
` + }, + { + text: `Решите уравнение $(x^2 + 4x + 5)^2 - 16(x^2 + 4x + 5) = 17$. + В ответ запишите целые корни уравнения, + удовлетворяющие неравенству $|x| \\leq 3$.`, + sol: `Метод замены переменной: если в уравнении повторяется одно и то же выражение, удобно обозначить его новой буквой и решить как обычное квадратное уравнение. +
Теорема Виета (обратная) для $t^2+pt+q=0$: $t_1+t_2=-p$, $t_1\\cdot t_2=q$. +
Дискриминант: $D=b^2-4ac$; при $D\\lt 0$ корней нет. +

+Шаг 1. Замечаем, что выражение $x^2+4x+5$ встречается дважды. Сделаем замену: +$$t = x^2+4x+5$$ +Тогда уравнение примет вид: +$$t^2 - 16t - 17 = 0$$ +Шаг 2. По теореме Виета: $t_1+t_2=16$, $t_1\\cdot t_2=-17$. Подходят $17$ и $-1$: +$$(t-17)(t+1)=0 \\implies t=17 \\text{ или } t=-1$$ +Шаг 3. Возвращаемся к переменной $x$. Рассмотрим два случая. +
Случай 1: $x^2+4x+5=17$, то есть $x^2+4x-12=0$. +
Раскладываем: $(x+6)(x-2)=0 \\Rightarrow x=-6$ или $x=2$. +
Проверяем условие $|x|\\leq 3$: $x=2$ подходит, $x=-6$ — нет. +
Случай 2: $x^2+4x+5=-1$, то есть $x^2+4x+6=0$. +
Дискриминант: $D=16-24=-8\\lt 0$, значит вещественных корней нет. +
Шаг 4. Из всех найденных корней условию $|x|\\leq 3$ удовлетворяет только $x=2$. +
Ответ: $x=2$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v04.js b/frontend/js/exam9/variants/v04.js new file mode 100644 index 0000000..8e64df6 --- /dev/null +++ b/frontend/js/exam9/variants/v04.js @@ -0,0 +1,216 @@ +VARIANTS[4] = { + label: "Вариант 4", + tasks: [ + { + text: `Определите, какое из следующих равенств верно:`, + opts: [ + ["а", "$b^2 \\cdot b^9 = b^{18}$"], ["б", "$b^2 \\cdot b^9 = 11b$"], + ["в", "$b^2 \\cdot b^9 = b^{11}$"], ["г", "$b^2 \\cdot b^9 = b^7$"], + ["д", "$b^2 \\cdot b^9 = 18b$"], + ], + sol: `При умножении степеней с одним основанием показатели складываются: +$$b^2 \\cdot b^9 = b^{2+9} = b^{11}$$ +
Ответ: в) $b^{11}$
` + }, + { + text: `Значение выражения $14\\dfrac{10}{21} - 2\\dfrac{12}{21}$ равно:`, + opts: [ + ["а", "$12\\dfrac{2}{21}$"], ["б", "$12\\dfrac{19}{21}$"], + ["в", "$11\\dfrac{19}{21}$"], ["г", "$10\\dfrac{19}{21}$"], + ["д", "$11\\dfrac{2}{21}$"], + ], + sol: `$$14\\tfrac{10}{21} - 2\\tfrac{12}{21} = (14-2) + \\left(\\tfrac{10}{21}-\\tfrac{12}{21}\\right) = 12 - \\tfrac{2}{21} = 11\\tfrac{21}{21} - \\tfrac{2}{21} = 11\\tfrac{19}{21}$$ +
Ответ: в) $11\\dfrac{19}{21}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "на плоскости через точку можно провести только одну прямую, параллельную данной;"], + ["б", "если в треугольнике два угла равны, то он равнобедренный;"], + ["в", "у любого прямоугольника диагонали перпендикулярны;"], + ["г", "сумма двух соседних углов параллелограмма равна $180^{\\circ}$?"], + ], + sol: `
    +
  • а) Через точку на плоскости проходит ровно одна прямая, параллельная данной — верно (аксиома параллельных)
    • +
  • б) Два равных угла ⟹ равнобедренный треугольник — верно
    • +
  • в) У любого прямоугольника диагонали перпендикулярны — НЕВЕРНО
    • +
  • г) Сумма двух соседних углов параллелограмма $= 180°$ — верно
    • +
+Диагонали прямоугольника равны между собой, но не обязательно перпендикулярны — перпендикулярны диагонали ромба. +
Ответ: в)
` + }, + { + text: `Определите количество целых решений системы неравенств + $$\\begin{cases} x < 8, \\\\[4pt] 5 - x \\leq 0. \\end{cases}$$`, + sol: `Из второго неравенства: $5-x \\leq 0 \\Rightarrow x \\geq 5$. +
Система: $5 \\leq x \\lt 8$ +
Целые числа: $5,\\ 6,\\ 7$ — ровно 3 числа. +
Ответ: 3
` + }, + { + text: `В окружность с радиусом $25$ см вписан треугольник, одна сторона которого + является диаметром, а другая — равна $14$ см. Найдите площадь этого треугольника.`, + sol: `Теорема Фалеса (о вписанном угле): вписанный угол, опирающийся на диаметр окружности, равен $90°$. +
Теорема Пифагора: в прямоугольном треугольнике $c^2 = a^2 + b^2$, где $c$ — гипотенуза. +
Формула площади прямоугольного треугольника: $S = \\dfrac{1}{2}\\cdot a\\cdot b$. +

+Шаг 1. Одна сторона треугольника — диаметр окружности, значит: +$$d = 2R = 2\\cdot 25 = 50\\text{ см}$$ +Шаг 2. По теореме Фалеса вписанный угол, опирающийся на диаметр, прямой ($90°$). Значит, треугольник прямоугольный, его гипотенуза $= 50$ см. + + + + +A +B +C +50 см +14 см + +Шаг 3. По условию один катет равен $14$ см. По теореме Пифагора находим второй катет: +$$b = \\sqrt{50^2-14^2} = \\sqrt{2500-196} = \\sqrt{2304} = 48\\text{ см}$$ +Шаг 4. Площадь прямоугольного треугольника — полупроизведение катетов: +$$S = \\tfrac{1}{2}\\cdot 14\\cdot 48 = 336\\text{ см}^2$$ +
Ответ: $336$ см²
` + }, + { + text: `Найдите значение выражения $\\dfrac{b - 5}{b - 2\\sqrt{5b} + 5}$ при $b = 20$.`, + sol: `Формула разности квадратов: $a^2-b^2 = (a-b)(a+b)$. +
Формула квадрата разности: $(a-b)^2 = a^2-2ab+b^2$. +
Идея: при работе с радикалами представляем $b$ как $(\\sqrt{b})^2$, чтобы увидеть структуру формул сокращённого умножения. +

+Шаг 1. Разложим знаменатель. Представим $b=(\\sqrt{b})^2$ и $5=(\\sqrt{5})^2$: +$$b - 2\\sqrt{5b} + 5 = (\\sqrt{b})^2 - 2\\sqrt{b}\\cdot\\sqrt{5} + (\\sqrt{5})^2 = (\\sqrt{b}-\\sqrt{5})^2$$ +(это квадрат разности). +
Шаг 2. Разложим числитель по формуле разности квадратов: +$$b - 5 = (\\sqrt{b})^2 - (\\sqrt{5})^2 = (\\sqrt{b}-\\sqrt{5})(\\sqrt{b}+\\sqrt{5})$$ +Шаг 3. Сокращаем общий множитель $(\\sqrt{b}-\\sqrt{5})$: +$$\\dfrac{(\\sqrt{b}-\\sqrt{5})(\\sqrt{b}+\\sqrt{5})}{(\\sqrt{b}-\\sqrt{5})^2} = \\dfrac{\\sqrt{b}+\\sqrt{5}}{\\sqrt{b}-\\sqrt{5}}$$ +Шаг 4. Подставляем $b=20$. Заметим, что $\\sqrt{20} = \\sqrt{4\\cdot 5} = 2\\sqrt{5}$: +$$\\dfrac{\\sqrt{20}+\\sqrt{5}}{\\sqrt{20}-\\sqrt{5}} = \\dfrac{2\\sqrt{5}+\\sqrt{5}}{2\\sqrt{5}-\\sqrt{5}} = \\dfrac{3\\sqrt{5}}{\\sqrt{5}} = 3$$ +
Ответ: $3$
` + }, + { + text: `Функция задана формулой $f(x) = \\dfrac{3}{x}$. + Расположите в порядке возрастания $f(-2{,}3)$, $f(-11{,}5)$, $f(-\\sqrt{3})$. + Ответ обоснуйте.`, + sol: `Свойство гиперболы $y=\\dfrac{k}{x}$ при $k\\gt 0$: функция убывает на каждом из промежутков $(-\\infty;0)$ и $(0;+\\infty)$. +
Правило сравнения значений убывающей функции: чем больше аргумент, тем меньше значение функции. +

+Шаг 1. Определим характер функции. У функции $f(x)=\\dfrac{3}{x}$ коэффициент $k=3\\gt 0$, поэтому на промежутке $(-\\infty;0)$ функция убывает. +
Шаг 2. Все три аргумента отрицательны, значит, лежат на промежутке убывания. +
Оценим $\\sqrt{3}$: так как $1\\lt 3\\lt 4$, то $1\\lt\\sqrt{3}\\lt 2$, точнее $\\sqrt{3}\\approx 1{,}73$, поэтому $-\\sqrt{3}\\approx -1{,}73$. +
Расставим аргументы по возрастанию: +$$-11{,}5 \\lt -2{,}3 \\lt -\\sqrt{3}$$ +Шаг 3. Так как функция убывает, при увеличении аргумента значение $f$ уменьшается: +$$f(-11{,}5) \\gt f(-2{,}3) \\gt f(-\\sqrt{3})$$ +Шаг 4. Перепишем по возрастанию (от меньшего к большему): +$$f(-\\sqrt{3}) \\lt f(-2{,}3) \\lt f(-11{,}5)$$ +Проверка вычислением: $f(-\\sqrt{3})=-\\sqrt{3}\\approx -1{,}73$; $f(-2{,}3)\\approx -1{,}30$; $f(-11{,}5)\\approx -0{,}26$. Действительно, $-1{,}73\\lt -1{,}30\\lt -0{,}26$ ✓. +
Ответ (по возрастанию): $f(-\\sqrt{3}) \\lt f(-2{,}3) \\lt f(-11{,}5)$
` + }, + { + text: `Найдите сумму всех двузначных натуральных чисел, + которые при делении на $11$ дают в остатке $6$.`, + sol: `Теорема о делении с остатком: если число $n$ при делении на $d$ даёт остаток $r$, то $n = d\\cdot k + r$, где $k$ — целое неотрицательное число. +
Формула суммы арифметической прогрессии: $S_n = \\dfrac{(a_1+a_n)\\cdot n}{2}$. +

+Шаг 1. Запишем общий вид искомых чисел. По условию число делится на $11$ с остатком $6$, поэтому имеет вид: +$$n = 11k + 6,\\quad k = 0, 1, 2, \\ldots$$ +Шаг 2. Найдём, при каких $k$ число будет двузначным: $10\\leq 11k+6\\leq 99$. +
Левое неравенство: $11k\\geq 4 \\Rightarrow k\\geq 1$. +
Правое неравенство: $11k\\leq 93 \\Rightarrow k\\leq 8$ (так как $11\\cdot 8=88\\leq 93$, а $11\\cdot 9=99\\gt 93$). +
Итак, $k = 1, 2, \\ldots, 8$ — всего $8$ значений. +
Шаг 3. Выпишем все двузначные числа: +$$17,\\ 28,\\ 39,\\ 50,\\ 61,\\ 72,\\ 83,\\ 94$$ +Это арифметическая прогрессия с $a_1=17$, разностью $d=11$, $n=8$, $a_8=94$. +
Шаг 4. По формуле суммы арифметической прогрессии: +$$S = \\dfrac{(a_1+a_n)\\cdot n}{2} = \\dfrac{(17+94)\\cdot 8}{2} = \\dfrac{111\\cdot 8}{2} = 111\\cdot 4 = 444$$ +
Ответ: $444$
` + }, + { + text: `Внутри угла $B$, равного $60^{\\circ}$, взята точка $K$. + Расстояния от точки $K$ до сторон угла равны $2$ см и $3$ см. + Найдите расстояние от точки $K$ до вершины угла $B$.`, + sol: `Опустим перпендикуляры $KH_1=2$ и $KH_2=3$ на стороны угла. + + + + + + d=BK + + + + + + + + 60° + + B + + + K + + + + + 2 + H₁ + + + + + 3 + H₂ + + + √19 + + + 120° + + + O + +Шаг 1 — угол при K.
+Четырёхугольник $BH_1KH_2$ имеет три известных угла. Сумма углов четырёхугольника $= 360°$: +$$\\angle H_1KH_2 = 360° - 90° - 60° - 90° = 120°$$ +Шаг 2 — длина отрезка H₁H₂.
+Теорема косинусов в $\\triangle H_1KH_2$: +$$H_1H_2^2 = 2^2 + 3^2 - 2\\cdot2\\cdot3\\cdot\\cos120° = 4+9-12\\cdot\\left(-\\tfrac{1}{2}\\right) = 13+6 = 19$$ +$$H_1H_2 = \\sqrt{19}$$ +Шаг 3 — четыре точки на одной окружности.
+Так как $\\angle BH_1K = \\angle BH_2K = 90°$, по обратной теореме Фалеса все четыре точки $B, H_1, K, H_2$ лежат на одной окружности с диаметром $BK$. +

+Шаг 4 — теорема синусов.
+В описанной окружности (диаметр $= BK$) вписанный угол $\\angle H_1BH_2 = 60°$ опирается на хорду $H_1H_2$: +$$\\dfrac{H_1H_2}{\\sin\\angle H_1BH_2} = BK \\implies BK = \\dfrac{\\sqrt{19}}{\\sin 60°} = \\dfrac{\\sqrt{19}}{\\tfrac{\\sqrt{3}}{2}} = \\dfrac{2\\sqrt{19}}{\\sqrt{3}} = \\dfrac{2\\sqrt{57}}{3}$$ +
Ответ: $\\dfrac{2\\sqrt{57}}{3}$ см
` + }, + { + text: `Решите уравнение $(x^2 + 2x + 3)^2 - 17(x^2 + 2x + 3) = 18$. + В ответ запишите целые корни уравнения, + удовлетворяющие неравенству $|x| \\leq 4$.`, + sol: `Метод замены переменной: если в уравнении повторяется одно и то же выражение, обозначаем его новой буквой и сводим к квадратному уравнению. +
Теорема Виета (обратная) для $t^2+pt+q=0$: $t_1+t_2=-p$, $t_1\\cdot t_2=q$. +
Дискриминант: $D=b^2-4ac$; при $D\\lt 0$ вещественных корней нет. +

+Шаг 1. Выражение $x^2+2x+3$ встречается дважды. Сделаем замену: +$$t = x^2+2x+3$$ +Уравнение примет вид: +$$t^2 - 17t - 18 = 0$$ +Шаг 2. По теореме Виета: $t_1+t_2=17$, $t_1\\cdot t_2=-18$. Подходят $18$ и $-1$: +$$(t-18)(t+1)=0 \\implies t=18 \\text{ или } t=-1$$ +Шаг 3. Возвращаемся к $x$. +
Случай 1: $x^2+2x+3=18$, то есть $x^2+2x-15=0$. +
Раскладываем: $(x+5)(x-3)=0 \\Rightarrow x=-5$ или $x=3$. +
Проверяем условие $|x|\\leq 4$: $x=3$ подходит, $x=-5$ — нет. +
Случай 2: $x^2+2x+3=-1$, то есть $x^2+2x+4=0$. +
Дискриминант: $D=4-16=-12\\lt 0$, значит корней нет. +
Шаг 4. Условию $|x|\\leq 4$ удовлетворяет только $x=3$. +
Ответ: $x=3$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v05.js b/frontend/js/exam9/variants/v05.js new file mode 100644 index 0000000..be74c58 --- /dev/null +++ b/frontend/js/exam9/variants/v05.js @@ -0,0 +1,205 @@ +VARIANTS[5] = { + label: "Вариант 5", + tasks: [ + { + text: `Определите наименьшее целое число, принадлежащее промежутку $[-8{,}5;\\, 3{,}4]$:`, + opts: [ + ["а", "$-9$"], ["б", "$-8$"], ["в", "$0$"], ["г", "$3$"], ["д", "$-1$"], + ], + sol: `Промежуток $[-8{,}5;\\, 3{,}4]$ включает $-8{,}5$ (закрытый конец). +
Ближайшее целое $\\geq -8{,}5$ — это $-8$. +
Ответ: б) $-8$
` + }, + { + text: `Второй член арифметической прогрессии $(a_n)$, + у которой $d = 2$ и $a_1 = \\dfrac{1}{2}$, равен:`, + opts: [ + ["а", "$1$"], ["б", "$1\\dfrac{1}{2}$"], ["в", "$2\\dfrac{1}{2}$"], + ["г", "$2$"], ["д", "$-1\\dfrac{1}{2}$"], + ], + sol: `$$a_2 = a_1 + d = \\dfrac{1}{2} + 2 = \\dfrac{5}{2} = 2\\dfrac{1}{2}$$ +
Ответ: в) $2\\dfrac{1}{2}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "сумма острых углов прямоугольного треугольника равна $90^{\\circ}$;"], + ["б", "диагонали параллелограмма точкой пересечения делятся пополам;"], + ["в", "любая точка серединного перпендикуляра к отрезку равноудалена от концов отрезка;"], + ["г", "если три угла одного треугольника соответственно равны трём углам другого, то такие треугольники равны?"], + ], + sol: `
    +
  • а) Сумма острых углов прямоугольного треугольника $= 90°$ — верно
    • +
  • б) Диагонали параллелограмма делятся пополам — верно
    • +
  • в) Точки серединного перпендикуляра равноудалены от концов — верно
    • +
  • г) Три равных угла → равные треугольники — НЕВЕРНО
    • +
+Три равных угла означают лишь подобие треугольников, но не равенство (они могут иметь разные стороны). +
Ответ: г)
` + }, + { + text: `Решите уравнение $2x^2 - 0{,}4x = 0$. + В ответ запишите среднее арифметическое корней уравнения.`, + sol: `Выносим $x$ за скобку: +$$x(2x - 0{,}4) = 0 \\implies x_1=0,\\quad x_2 = \\frac{0{,}4}{2} = 0{,}2$$ +Среднее арифметическое: $\\dfrac{x_1+x_2}{2} = \\dfrac{0+0{,}2}{2} = 0{,}1$ +
Ответ: $0{,}1$
` + }, + { + text: `Высоты параллелограмма, проведённые из вершины тупого угла, равны $6$ см и $9$ см. + Периметр параллелограмма равен $40$ см. Найдите площадь параллелограмма.`, + figure: ` + + + тупой + A + B + C + D + a + b + + + h₁=6 + + + h₂=9 +`, + sol: `Пусть стороны параллелограмма $a$ и $b$. Высоты из вершины тупого угла перпендикулярны к смежным сторонам: $h_a=6$ (к стороне $a$) и $h_b=9$ (к стороне $b$). +
Площадь одна и та же: $S = a\\cdot h_a = b\\cdot h_b$, значит: +$$6a = 9b \\implies \\frac{a}{b} = \\frac{3}{2}$$ +Периметр: $2(a+b)=40 \\Rightarrow a+b=20$. +
С учётом $a=3k,\\ b=2k$: $5k=20 \\Rightarrow k=4$, т.е. $a=12$, $b=8$. +$$S = a\\cdot h_a = 12\\cdot 6 = 72\\text{ см}^2$$ +
Ответ: $72$ см²
` + }, + { + text: `При каких натуральных значениях $n$ верно неравенство + $4{,}8(n - 4) - 3{,}7(2 - n) < 24{,}4$?`, + sol: `Свойства линейных неравенств: можно прибавлять одинаковое число к обеим частям и умножать/делить на положительное число, не меняя знак неравенства. +
Натуральные числа: $1, 2, 3, \\ldots$ — положительные целые. +

+Шаг 1. Раскроем скобки слева: +$$4{,}8(n-4) = 4{,}8n - 19{,}2$$ +$$-3{,}7(2-n) = -7{,}4 + 3{,}7n$$ +Неравенство примет вид: +$$4{,}8n - 19{,}2 - 7{,}4 + 3{,}7n \\lt 24{,}4$$ +Шаг 2. Приведём подобные слагаемые: +$$8{,}5n - 26{,}6 \\lt 24{,}4$$ +Шаг 3. Перенесём $-26{,}6$ вправо со сменой знака: +$$8{,}5n \\lt 51$$ +Шаг 4. Разделим обе части на $8{,}5$ (положительное число, знак сохраняется): +$$n \\lt 6$$ +Шаг 5. Выбираем натуральные значения, удовлетворяющие $n\\lt 6$: +$$n \\in \\{1,\\ 2,\\ 3,\\ 4,\\ 5\\}$$ +
Ответ: $n = 1,\\ 2,\\ 3,\\ 4,\\ 5$
` + }, + { + text: `Известно, что график функции $y = f(x)$ симметричен относительно оси ординат + и $f(-3) = 5$, $f(2) = -6$. + Найдите значение выражения $f(3) + 2f(-2)$.`, + sol: `Признак чётной функции: график функции симметричен относительно оси ординат тогда и только тогда, когда функция чётная. +
Свойство чётной функции: $f(-x) = f(x)$ для всех $x$ из области определения. +

+Шаг 1. По признаку чётной функции из симметрии графика относительно оси $Oy$ следует: +$$f(-x) = f(x)$$ +Шаг 2. Применяем это свойство, чтобы выразить нужные значения через известные. +
Так как $f(3) = f(-3)$, то по условию: +$$f(3) = f(-3) = 5$$ +Шаг 3. Аналогично для $f(-2)$: +$$f(-2) = f(2) = -6$$ +Шаг 4. Подставляем найденные значения в выражение: +$$f(3) + 2f(-2) = 5 + 2\\cdot(-6) = 5 - 12 = -7$$ +
Ответ: $-7$
` + }, + { + text: `Определите число решений системы уравнений + $$\\begin{cases} x^2 + y^2 = 16, \\\\[4pt] y = -x^2 + 4. \\end{cases}$$ + Ответ обоснуйте.`, + sol: `Метод подстановки: подставляем выражение для $y$ из одного уравнения в другое. +
Формула квадрата суммы: $(a+b)^2 = a^2+2ab+b^2$. +
Геометрический смысл: первое уравнение задаёт окружность с центром в начале координат и радиусом $4$; второе — параболу с вершиной $(0;4)$, ветви вниз. +

+Шаг 1. Из второго уравнения возьмём выражение $y=-x^2+4$ и подставим в первое: +$$x^2+(-x^2+4)^2=16$$ +Шаг 2. Раскрываем квадрат: +$$x^2+x^4-8x^2+16=16$$ +Шаг 3. Приводим подобные и упрощаем: +$$x^4-7x^2=0$$ +$$x^2(x^2-7)=0$$ +Шаг 4. Произведение равно нулю, когда один из множителей нуль: +
    +
  • $x^2=0 \\Rightarrow x=0$
    • +
  • $x^2-7=0 \\Rightarrow x=\\pm\\sqrt{7}$
    • +
+Получили три значения $x$. Поскольку $y$ однозначно определяется как $y=-x^2+4$, каждому значению $x$ соответствует одно значение $y$: +
$x=0$:$y=4$— точка $(0,4)$
$x=\\pm\\sqrt{7}$:$y=-3$— точки $(\\pm\\sqrt{7},\\,-3)$
+Шаг 5. Итого — три решения системы: + + + + +xy + +x²+y²=16 + +y=−x²+4 +(0,4) +(√7,−3) +(−√7,−3) + +
Ответ: 3 решения
` + }, + { + text: `К раствору, содержащему $30$ г соли, добавили $100$ г воды, + после чего концентрация соли уменьшилась на $5\\%$. + Найдите первоначальную процентную концентрацию соли в растворе.`, + sol: `Пусть $m$ — начальная масса раствора. Соли — $30$ г, она не меняется. +

+ + + + +
Масса раствораМасса солиКонцентрация
До$m$$30$ г$c = \\dfrac{30}{m}$
После$m+100$$30$ г$c' = \\dfrac{30}{m+100}$
+
+Шаг 1. По условию концентрация уменьшилась на $5\\% = 0{,}05$: +$$\\frac{30}{m} - \\frac{30}{m+100} = 0{,}05$$ +Шаг 2. Приведём к общему знаменателю: +$$\\frac{30(m+100) - 30m}{m(m+100)} = 0{,}05 \\implies \\frac{3000}{m(m+100)} = \\frac{1}{20}$$ +$$m(m+100) = 60000$$ +Шаг 3. Решаем квадратное уравнение $m^2 + 100m - 60000 = 0$: +$$D = 100^2 + 4\\cdot60000 = 10000 + 240000 = 250000 = 500^2$$ +$$m = \\frac{-100 + 500}{2} = 200\\text{ г} \\quad (m > 0)$$ +Шаг 4. Начальная концентрация: +$$c = \\frac{30}{200} = 0{,}15 = 15\\%$$ +Проверка: после добавления $c' = \\dfrac{30}{300} = 10\\% = 15\\% - 5\\%$ ✓ +
Ответ: $15\\%$
` + }, + { + text: `Найдите площадь сектора круга, угол которого равен $30^{\\circ}$, + а длина дуги — $4$ см. Ответ округлите до целых см², взяв $\\pi \\approx 3{,}14$.`, + sol: `Формула длины дуги: $l = \\dfrac{\\pi r\\alpha°}{180°}$, где $\\alpha°$ — центральный угол сектора в градусах. +
Формула площади сектора через длину дуги: $S = \\dfrac{l\\cdot r}{2}$. +

+Шаг 1. По формуле длины дуги при $l=4$ и $\\alpha=30°$ найдём радиус: +$$4 = \\dfrac{\\pi r\\cdot 30}{180} = \\dfrac{\\pi r}{6}$$ +Отсюда: +$$r = \\dfrac{24}{\\pi}$$ +Шаг 2. Подставляем найденный радиус в формулу площади сектора: +$$S = \\dfrac{l\\cdot r}{2} = \\dfrac{4\\cdot\\dfrac{24}{\\pi}}{2} = \\dfrac{48}{\\pi}$$ +Шаг 3. Подставляем $\\pi\\approx 3{,}14$ и округляем: +$$S \\approx \\dfrac{48}{3{,}14} \\approx 15{,}3 \\approx 15\\text{ см}^2$$ + + + + + + +30° +r +l=4 + +
Ответ: $15$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v06.js b/frontend/js/exam9/variants/v06.js new file mode 100644 index 0000000..ba4fd0d --- /dev/null +++ b/frontend/js/exam9/variants/v06.js @@ -0,0 +1,194 @@ +VARIANTS[6] = { + label: "Вариант 6", + tasks: [ + { + text: `Определите наибольшее целое число, принадлежащее промежутку $[-10{,}5;\\, -1{,}4]$:`, + opts: [ + ["а", "$-11$"], ["б", "$-10$"], ["в", "$-1$"], ["г", "$-2$"], ["д", "$0$"], + ], + sol: `Промежуток $[-10{,}5;\\, -1{,}4]$ имеет правый конец $-1{,}4$ (открытый). +
Наибольшее целое $\\leq -1{,}4$ — это $-2$. +
Ответ: г) $-2$
` + }, + { + text: `Второй член арифметической прогрессии $(a_n)$, + у которой $d = 1{,}5$ и $a_1 = -0{,}5$, равен:`, + opts: [ + ["а", "$1$"], ["б", "$1{,}5$"], ["в", "$-2$"], ["г", "$2$"], ["д", "$-0{,}75$"], + ], + sol: `$$a_2 = a_1 + d = -0{,}5 + 1{,}5 = 1$$ +
Ответ: а) $1$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "в равностороннем треугольнике все углы равны по $30^{\\circ}$;"], + ["б", "диагональ делит параллелограмм на два равных треугольника;"], + ["в", "любая точка биссектрисы угла равноудалена от сторон угла;"], + ["г", "если три стороны одного треугольника соответственно равны трём сторонам другого, то такие треугольники равны?"], + ], + sol: `
  • а) В равностороннем треугольнике все углы равны по $30^\\circ$ — НЕВЕРНО
  • б) Диагональ делит параллелограмм на два равных треугольника — верно
  • в) Точки биссектрисы угла равноудалены от сторон угла — верно
  • г) Три равных стороны → равные треугольники — верно
В равностороннем треугольнике каждый угол $= 60^\\circ$, а не $30^\\circ$.
Ответ: а)
` + }, + { + text: `Решите уравнение $3x^2 + 0{,}6x = 0$. + В ответ запишите среднее арифметическое корней уравнения.`, + sol: `Выносим $x$ за скобку: +$$x(3x + 0{,}6) = 0 \\implies x_1=0,\\quad x_2 = -\\frac{0{,}6}{3} = -0{,}2$$ +Среднее арифметическое: $\\dfrac{x_1+x_2}{2} = \\dfrac{0+(-0{,}2)}{2} = -0{,}1$ +
Ответ: $-0{,}1$
` + }, + { + text: `Высоты параллелограмма, проведённые из вершины тупого угла, равны $10$ см и $6$ см. + Периметр параллелограмма равен $48$ см. Найдите площадь параллелограмма.`, + sol: ` + + + тупой + A + B + C + D + a + b + + + h₁=10 + + + h₂=6 + +Пусть стороны параллелограмма $a$ и $b$. Высоты из вершины тупого угла перпендикулярны к смежным сторонам: $h_a=10$ (к стороне $a$) и $h_b=6$ (к стороне $b$). +
Площадь одна и та же: $S = a\\cdot h_a = b\\cdot h_b$, значит: +$$10a = 6b \\implies \\frac{a}{b} = \\frac{3}{5}$$ +Периметр: $2(a+b)=48 \\Rightarrow a+b=24$. +
С учётом $a=3k,\\ b=5k$: $8k=24 \\Rightarrow k=3$, т.е. $a=9$, $b=15$. +$$S = a\\cdot h_a = 9\\cdot 10 = 90\\text{ см}^2$$ +
Ответ: $90$ см²
` + }, + { + text: `При каких натуральных значениях $n$ верно неравенство + $5{,}6(n - 3) - 3{,}2(2 - n) < 20{,}8$?`, + sol: `Свойства линейных неравенств: можно прибавлять к обеим частям одно и то же число и делить на положительное число — знак неравенства сохраняется. +
Натуральные числа: $1, 2, 3, \\ldots$ +

+Шаг 1. Раскроем скобки в левой части: +$$5{,}6(n-3) = 5{,}6n - 16{,}8$$ +$$-3{,}2(2-n) = -6{,}4 + 3{,}2n$$ +Неравенство: +$$5{,}6n - 16{,}8 - 6{,}4 + 3{,}2n \\lt 20{,}8$$ +Шаг 2. Приведём подобные: +$$8{,}8n - 23{,}2 \\lt 20{,}8$$ +Шаг 3. Перенесём $-23{,}2$ направо: +$$8{,}8n \\lt 44$$ +Шаг 4. Разделим на $8{,}8$ (положительное число — знак сохраняется): +$$n \\lt 5$$ +Шаг 5. Натуральные значения, удовлетворяющие $n\\lt 5$: +$$n \\in \\{1,\\ 2,\\ 3,\\ 4\\}$$ +
Ответ: $n = 1,\\ 2,\\ 3,\\ 4$
` + }, + { + text: `Известно, что график функции $y = f(x)$ симметричен относительно оси ординат + и $f(-5) = 2$, $f(4) = -5$. + Найдите значение выражения $2f(5) - f(-4)$.`, + sol: `Признак чётной функции: график функции симметричен относительно оси ординат тогда и только тогда, когда функция чётная. +
Свойство чётной функции: $f(-x) = f(x)$ для всех $x$ из области определения. +

+Шаг 1. Из симметрии графика относительно оси $Oy$ следует, что функция чётная, то есть: +$$f(-x) = f(x)$$ +Шаг 2. Выразим $f(5)$ через известное значение $f(-5)$, используя $f(5)=f(-5)$: +$$f(5) = f(-5) = 2$$ +Шаг 3. Аналогично для $f(-4)$: +$$f(-4) = f(4) = -5$$ +Шаг 4. Подставляем найденные значения: +$$2f(5) - f(-4) = 2\\cdot 2 - (-5) = 4 + 5 = 9$$ +
Ответ: $9$
` + }, + { + text: `Определите число решений системы уравнений + $$\\begin{cases} x^2 + y^2 = 9, \\\\[4pt] y = -x^2 + 3. \\end{cases}$$ + Ответ обоснуйте.`, + sol: `Метод подстановки для системы уравнений: подставляем выражение для $y$ из одного уравнения в другое. +
Геометрический смысл: первое уравнение — окружность радиуса $3$ с центром в начале координат; второе — парабола с вершиной $(0;3)$ и ветвями вниз. +

+Шаг 1. Из второго уравнения возьмём $y=-x^2+3$ и подставим в первое: +$$x^2+(-x^2+3)^2=9$$ +Шаг 2. Раскрываем квадрат и приводим подобные: +$$x^2 + x^4 - 6x^2 + 9 = 9$$ +$$x^4 - 5x^2 = 0$$ +Шаг 3. Вынесем общий множитель: +$$x^2(x^2-5)=0$$ +Произведение равно нулю, когда один из множителей нуль: $x=0$ или $x^2=5$, то есть $x=\\pm\\sqrt{5}$. +
Шаг 4. Для каждого $x$ находим $y$ по формуле $y=-x^2+3$: +
    +
  • $x=0 \\Rightarrow y=3$ — точка $(0;3)$;
    • +
  • $x=\\pm\\sqrt{5} \\Rightarrow y=-5+3=-2$ — точки $(\\pm\\sqrt{5};-2)$.
    • +
+Шаг 5. Получили три точки пересечения, значит, у системы 3 решения. + + + + +xy + +x²+y²=9 + +y=−x²+3 +(0,3) +(√5,−2) +(−√5,−2) + +
Ответ: 3 решения
` + }, + { + text: `К раствору, содержащему $40$ г соли, добавили $200$ г воды, + после чего концентрация соли уменьшилась на $10\\%$. + Найдите первоначальную процентную концентрацию соли в растворе.`, + sol: `Формула концентрации: $c = \\dfrac{m_{\\text{соли}}}{m_{\\text{раствора}}}$ — отношение массы соли к массе всего раствора. +
Ключевое наблюдение: при добавлении воды масса соли не меняется, изменяется только масса всего раствора и, следовательно, концентрация. +

+Шаг 1. Введём переменную: пусть $m$ (г) — начальная масса раствора. +
В нём $40$ г соли. После добавления $200$ г воды масса раствора стала $m+200$ г, а соли по-прежнему $40$ г. +
Шаг 2. Запишем концентрации до и после: +$$c_{\\text{до}} = \\dfrac{40}{m},\\qquad c_{\\text{после}} = \\dfrac{40}{m+200}$$ +Шаг 3. По условию концентрация уменьшилась на $10\\%$, то есть $c_{\\text{до}} - c_{\\text{после}} = 0{,}1$: +$$\\dfrac{40}{m} - \\dfrac{40}{m+200} = 0{,}1$$ +Шаг 4. Приводим к общему знаменателю $m(m+200)$: +$$\\dfrac{40(m+200) - 40m}{m(m+200)} = 0{,}1$$ +$$\\dfrac{8000}{m(m+200)} = 0{,}1$$ +$$m(m+200) = 80000$$ +Шаг 5. Раскрываем скобки и получаем квадратное уравнение: +$$m^2 + 200m - 80000 = 0$$ +Дискриминант: $D = 200^2 + 4\\cdot 80000 = 40000 + 320000 = 360000 = 600^2$. +$$m = \\dfrac{-200 + 600}{2} = 200\\text{ г}\\quad (m\\gt 0)$$ +Шаг 6. Находим начальную концентрацию: +$$c_{\\text{до}} = \\dfrac{40}{200} = 0{,}2 = 20\\%$$ +Проверка: после добавления $c_{\\text{после}} = \\dfrac{40}{400} = 10\\% = 20\\%-10\\%$ ✓ +
Ответ: $20\\%$
` + }, + { + text: `Найдите площадь сектора круга, угол которого равен $150^{\\circ}$, + а длина дуги — $6$ см. Ответ округлите до целых см², взяв $\\pi \\approx 3{,}14$.`, + sol: `Формула длины дуги: $l = \\dfrac{\\pi r\\alpha°}{180°}$. +
Формула площади сектора через длину дуги: $S = \\dfrac{l\\cdot r}{2}$. +

+Шаг 1. Из формулы длины дуги при $l=6$, $\\alpha=150°$ найдём радиус: +$$6 = \\dfrac{\\pi r\\cdot 150}{180} = \\dfrac{5\\pi r}{6}$$ +$$r = \\dfrac{36}{5\\pi}$$ +Шаг 2. Подставим в формулу площади: +$$S = \\dfrac{l\\cdot r}{2} = \\dfrac{6\\cdot\\dfrac{36}{5\\pi}}{2} = \\dfrac{108}{5\\pi}$$ +Шаг 3. Подставим $\\pi\\approx 3{,}14$: +$$S \\approx \\dfrac{108}{5\\cdot 3{,}14} = \\dfrac{108}{15{,}7} \\approx 6{,}88 \\approx 7\\text{ см}^2$$ + + + + + + +150° +r +l=6 + +
Ответ: $7$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v07.js b/frontend/js/exam9/variants/v07.js new file mode 100644 index 0000000..b085bd4 --- /dev/null +++ b/frontend/js/exam9/variants/v07.js @@ -0,0 +1,290 @@ +VARIANTS[7] = { + label: "Вариант 7", + tasks: [ + { + text: `Определите, какое из следующих равенств верно:`, + opts: [ + ["а", "$a^{-4} = -4a$"], + ["б", "$a^{-4} = -a^4$"], + ["в", "$a^{-4} = \\dfrac{1}{a^4}$"], + ["г", "$a^{-4} = -\\dfrac{4}{a}$"], + ["д", "$a^{-4} = -\\dfrac{1}{a^4}$"], + ], + sol: `По определению отрицательного показателя: $a^{-n} = \\dfrac{1}{a^n}$. +$$a^{-4} = \\frac{1}{a^4}$$ +Остальные варианты неверны: знаменатель $a^{-4}$ всегда положителен при $a\\neq 0$. +
Ответ: в) $a^{-4}=\\dfrac{1}{a^4}$
` + }, + { + text: `Второй член геометрической прогрессии $(b_n)$, + у которой $q = 3$ и $b_1 = \\dfrac{2}{3}$, равен:`, + opts: [ + ["а", "$1$"], ["б", "$2$"], ["в", "$\\dfrac{2}{9}$"], + ["г", "$-2\\dfrac{1}{3}$"], ["д", "$3\\dfrac{2}{3}$"], + ], + sol: `Каждый следующий член геометрической прогрессии умножается на знаменатель $q$: +$$b_2 = b_1\\cdot q = \\dfrac{2}{3}\\cdot 3 = 2$$ +
Ответ: б) $2$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "накрест лежащие углы при двух параллельных прямых и секущей равны между собой;"], + ["б", "средняя линия треугольника параллельна основанию;"], + ["в", "$\\sin 30^{\\circ} = \\dfrac{1}{2}$;"], + ["г", "если диагонали параллелограмма равны, то это обязательно квадрат?"], + ], + sol: `
    +
  • а) Накрест лежащие углы при ∥ прямых равны — верно
    • +
  • б) Средняя линия треугольника ∥ основанию — верно
    • +
  • в) $\\sin 30^\\circ = \\frac{1}{2}$ — верно
    • +
  • г) Равные диагонали → квадрат — НЕВЕРНО
    • +
+Если диагонали параллелограмма равны, он является прямоугольником, но не обязательно квадратом. Прямоугольник $3\\times 4$ имеет равные диагонали, но это не квадрат. +
Ответ: г)
` + }, + { + text: `Определите наименьшее целое решение двойного неравенства + $-2 < \\dfrac{3x + 1}{2} \\leq 5$.`, + sol: `Умножим все части на $2$: +$$-4 < 3x+1 \\leq 10$$ +Вычтем $1$: +$$-5 < 3x \\leq 9$$ +Разделим на $3$: +$$-\\dfrac{5}{3} < x \\leq 3$$ + + + + −2 + −1 + 0 + 1 + 2 + 3 + + + + −5/3 + +$x\\in\\left(-\\dfrac{5}{3};\\,3\\right]$. Наименьшее целое число, большее $-\\dfrac{5}{3}\\approx-1{,}67$: это $-1$. +
Ответ: $-1$
` + }, + { + text: `В трапеции $ABCD$ с основаниями $AD$ и $BC$ точка $O$ — пересечение диагоналей, + $AD = 10$ см, $AO = 6$ см, $OC = 3$ см. Найдите среднюю линию трапеции.`, + sol: `Свойство диагоналей трапеции: точка пересечения диагоналей делит каждую диагональ в одном и том же отношении, равном отношению оснований. +
Формула средней линии трапеции: $m = \\dfrac{AD+BC}{2}$ — полусумма оснований. +

+Шаг 1. Точка $O$ — пересечение диагоналей трапеции. По свойству: +$$\\dfrac{AO}{OC} = \\dfrac{AD}{BC}$$ +(основания $AD$ и $BC$, точка $O$ ближе к меньшему основанию). +
Шаг 2. Подставляем известные значения $AO=6$, $OC=3$, $AD=10$: +$$\\dfrac{6}{3} = \\dfrac{10}{BC}$$ +$$BC = \\dfrac{10\\cdot 3}{6} = 5\\text{ см}$$ + + + + + + A + D + C + B + O + AD = 10 + BC = 5 + AO=6 + OC=3 + +Шаг 3. Находим среднюю линию трапеции по формуле: +$$m = \\dfrac{AD + BC}{2} = \\dfrac{10+5}{2} = 7{,}5\\text{ см}$$ +
Ответ: $7{,}5$ см
` + }, + { + text: `Упростите выражение + $$\\dfrac{x - y}{\\sqrt{x} + \\sqrt{y}} - \\dfrac{x + 4\\sqrt{xy} + 4y}{\\sqrt{x} + 2\\sqrt{y}}.$$`, + sol: `Формула разности квадратов: $a^2-b^2 = (a-b)(a+b)$. +
Формула квадрата суммы: $(a+b)^2 = a^2+2ab+b^2$. +
Идея: представляем $x$ как $(\\sqrt{x})^2$, чтобы использовать формулы сокращённого умножения и сократить дроби. +

+Шаг 1. Преобразуем первую дробь. Числитель $x-y$ — разность квадратов: +$$x - y = (\\sqrt{x})^2 - (\\sqrt{y})^2 = (\\sqrt{x}-\\sqrt{y})(\\sqrt{x}+\\sqrt{y})$$ +Сокращаем общий множитель $(\\sqrt{x}+\\sqrt{y})$: +$$\\dfrac{x-y}{\\sqrt{x}+\\sqrt{y}} = \\dfrac{(\\sqrt{x}-\\sqrt{y})(\\sqrt{x}+\\sqrt{y})}{\\sqrt{x}+\\sqrt{y}} = \\sqrt{x}-\\sqrt{y}$$ +Шаг 2. Преобразуем вторую дробь. Числитель — квадрат суммы: +$$x + 4\\sqrt{xy} + 4y = (\\sqrt{x})^2 + 2\\cdot\\sqrt{x}\\cdot 2\\sqrt{y} + (2\\sqrt{y})^2 = (\\sqrt{x}+2\\sqrt{y})^2$$ +Сокращаем общий множитель $(\\sqrt{x}+2\\sqrt{y})$: +$$\\dfrac{(\\sqrt{x}+2\\sqrt{y})^2}{\\sqrt{x}+2\\sqrt{y}} = \\sqrt{x}+2\\sqrt{y}$$ +Шаг 3. Вычисляем разность преобразованных выражений: +$$(\\sqrt{x}-\\sqrt{y}) - (\\sqrt{x}+2\\sqrt{y}) = \\sqrt{x} - \\sqrt{y} - \\sqrt{x} - 2\\sqrt{y} = -3\\sqrt{y}$$ +
Ответ: $-3\\sqrt{y}$
` + }, + { + text: `График функции $y = g(x)$ получен из графика функции $f(x) = x^2$ + сдвигом на $1$ единицу вправо вдоль оси абсцисс и на $9$ единиц вниз + вдоль оси ординат. Найдите нули функции $y = g(x)$.`, + sol: `Правило сдвига графика функции: +
• сдвиг на $a$ единиц вправо по оси $Ox$: $f(x) \\to f(x-a)$; +
• сдвиг на $b$ единиц вниз по оси $Oy$: $f(x) \\to f(x)-b$. +
Нули функции: это значения $x$, при которых $f(x)=0$ (точки пересечения графика с осью $Ox$). +
Формула разности квадратов: $a^2-b^2=(a-b)(a+b)$. +

+Шаг 1. Применим первый сдвиг: $f(x)=x^2$ сдвигаем на $1$ единицу вправо. По правилу получим: +$$f_1(x) = (x-1)^2$$ +Шаг 2. Применим второй сдвиг: $f_1(x)$ сдвигаем на $9$ единиц вниз: +$$g(x) = (x-1)^2 - 9$$ +Шаг 3. Чтобы найти нули, решаем уравнение $g(x)=0$: +$$(x-1)^2 - 9 = 0$$ +$$(x-1)^2 = 9$$ +Шаг 4. Извлекаем квадратный корень из обеих частей: +$$x-1 = \\pm 3$$ +Значит, $x-1=3$ (тогда $x=4$) или $x-1=-3$ (тогда $x=-2$). + + + + + xy + + + −2 + + 4 + + (1; −9) + 0 + + 1 + +
Ответ: $x = -2$ и $x = 4$
` + }, + { + text: `Решите уравнение $\\dfrac{5}{x^2 - x - 6} + \\dfrac{1}{x + 2} = -1$.`, + sol: `Решение дробно-рациональных уравнений состоит из трёх шагов: +
1) найти ОДЗ — все значения переменной, при которых знаменатели не равны нулю; +
2) привести к общему знаменателю и упростить; +
3) проверить, входят ли найденные корни в ОДЗ. +
Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. Разложим знаменатель первой дроби. Ищем числа с суммой $1$ и произведением $-6$. Это $3$ и $-2$: +$$x^2-x-6 = (x-3)(x+2)$$ +Шаг 2. Запишем ОДЗ: знаменатели не должны быть равны нулю, поэтому +$$x\\neq 3,\\quad x\\neq -2$$ +Шаг 3. Умножим обе части уравнения на $(x-3)(x+2)$, чтобы избавиться от дробей: +$$5 + (x-3) = -(x-3)(x+2)$$ +Шаг 4. Упрощаем: +$$x+2 = -(x^2-x-6)$$ +$$x+2 = -x^2+x+6$$ +$$x^2 - 4 = 0$$ +Шаг 5. Решаем как разность квадратов: $(x-2)(x+2)=0 \\Rightarrow x=2$ или $x=-2$. +
Шаг 6. Проверяем ОДЗ: $x=-2$ не подходит (исключён). Проверяем $x=2$ подстановкой: +$$\\dfrac{5}{4-2-6}+\\dfrac{1}{4} = \\dfrac{5}{-4}+\\dfrac{1}{4} = -1 \\checkmark$$ +
Ответ: $x = 2$
` + }, + { + text: `Если двузначное число разделить на сумму его цифр, то в частном получится $7$ + и в остатке $6$. Если это же двузначное число разделить на произведение его цифр, + то в частном получится $3$ и в остатке $11$. Найдите это двузначное число.`, + sol: `Запись двузначного числа: любое двузначное число можно представить как $10a+b$, где $a$ — цифра десятков ($1\\leq a\\leq 9$), $b$ — цифра единиц ($0\\leq b\\leq 9$). +
Теорема о делении с остатком: если число $N$ при делении на $d$ даёт частное $q$ и остаток $r$, то $N = d\\cdot q + r$, причём $0\\leq r\\lt d$. +
Формула корней квадратного уравнения: $x=\\dfrac{-b\\pm\\sqrt{D}}{2a}$, где $D=b^2-4ac$. +

+Шаг 1. Обозначим за $10a+b$ искомое двузначное число (где $a$ — цифра десятков, $b$ — цифра единиц). +
Шаг 2. Запишем первое условие. Сумма цифр — это $a+b$. При делении $10a+b$ на $a+b$ получили частное $7$ и остаток $6$: +$$10a + b = 7(a+b) + 6$$ +$$10a + b = 7a + 7b + 6$$ +$$3a - 6b = 6$$ +$$a = 2b + 2 \\quad (*)$$ +Шаг 3. Запишем второе условие. Произведение цифр — $ab$. При делении на $ab$ получили частное $3$ и остаток $11$: +$$10a + b = 3ab + 11$$ +Шаг 4. Подставим выражение $(*)$ для $a$ во второе условие: +$$10(2b+2) + b = 3(2b+2)b + 11$$ +$$20b + 20 + b = 6b^2 + 6b + 11$$ +$$21b + 20 = 6b^2 + 6b + 11$$ +$$6b^2 - 15b - 9 = 0$$ +Делим на $3$: +$$2b^2 - 5b - 3 = 0$$ +Шаг 5. Решаем квадратное уравнение. Дискриминант: $D = 25 + 24 = 49 = 7^2$. +$$b = \\dfrac{5\\pm 7}{4} \\implies b = 3 \\text{ или } b = -\\dfrac{1}{2}$$ +Цифра должна быть целым числом от $0$ до $9$, поэтому подходит только $b=3$. +
Шаг 6. Из формулы $(*)$ находим $a$: +$$a = 2\\cdot 3 + 2 = 8$$ +Искомое число: $\\boldsymbol{10\\cdot 8 + 3 = 83}$. +
Проверка: +
• сумма цифр $= 8+3 = 11$; $83:11 = 7$ (ост. $6$): $7\\cdot 11+6 = 77+6=83$ ✓; +
• произведение цифр $= 8\\cdot 3 = 24$; $83:24 = 3$ (ост. $11$): $3\\cdot 24+11 = 72+11=83$ ✓. +
Ответ: $83$
` + }, + { + text: `Внутри параллелограмма $ABCD$ взята точка $M$, такая, что + $S_{BMC} = 6$ см², $S_{AMD} = 10$ см². + Найдите площадь параллелограмма $ABCD$.`, + figure: ` + + + + + + + + + A + B + C + D + M + 10 см² + 6 см² +`, + sol: `Доказательство ключевого свойства: +$$S_{\\triangle BMC} + S_{\\triangle AMD} = \\dfrac{S_{ABCD}}{2}$$ +Стороны $BC \\parallel DA$ и $|BC| = |DA| = a$ (как противоположные стороны параллелограмма). +
Пусть $h_1$ — расстояние от $M$ до стороны $BC$, $h_2$ — расстояние от $M$ до стороны $DA$. + + + + + + + + a + a + + + M + + + + + + + h₁ + + + + + + + h₂ + + + + + H + + A + B + C + D + + $S_1$ + $S_2$ + +Так как $BC \\parallel DA$, то $h_1 + h_2 = H$ (полное расстояние между параллельными сторонами). +$$S_{\\triangle BMC} = \\tfrac{1}{2}\\cdot a\\cdot h_1 \\qquad S_{\\triangle AMD} = \\tfrac{1}{2}\\cdot a\\cdot h_2$$ +$$S_{\\triangle BMC}+S_{\\triangle AMD} = \\tfrac{1}{2}\\cdot a\\cdot(h_1+h_2) = \\tfrac{1}{2}\\cdot a\\cdot H = \\boxed{\\dfrac{S_{ABCD}}{2}}$$ +Вычисление: +$$\\dfrac{S_{ABCD}}{2} = S_{\\triangle BMC}+S_{\\triangle AMD} = 6+10 = 16 \\implies S_{ABCD} = 32\\text{ см}^2$$ +
Ответ: $32$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v08.js b/frontend/js/exam9/variants/v08.js new file mode 100644 index 0000000..98481c5 --- /dev/null +++ b/frontend/js/exam9/variants/v08.js @@ -0,0 +1,233 @@ +VARIANTS[8] = { + label: "Вариант 8", + tasks: [ + { + text: `Определите, какое из следующих равенств верно:`, + opts: [ + ["а", "$b^{-3} = -3b$"], + ["б", "$b^{-3} = -b^3$"], + ["в", "$b^{-3} = \\dfrac{1}{b^3}$"], + ["г", "$b^{-3} = -\\dfrac{1}{b^3}$"], + ["д", "$b^{-3} = -\\dfrac{3}{b}$"], + ], + sol: `По определению отрицательного показателя: $b^{-n} = \\dfrac{1}{b^n}$. +$$b^{-3} = \\frac{1}{b^3}$$ +
Ответ: в) $b^{-3}=\\dfrac{1}{b^3}$
` + }, + { + text: `Второй член геометрической прогрессии $(b_n)$, + у которой $q = 4$ и $b_1 = \\dfrac{3}{8}$, равен:`, + opts: [ + ["а", "$1$"], ["б", "$2$"], ["в", "$\\dfrac{1}{2}$"], + ["г", "$\\dfrac{2}{3}$"], ["д", "$1\\dfrac{1}{2}$"], + ], + sol: `$$b_2 = b_1\\cdot q = \\dfrac{3}{8}\\cdot 4 = \\dfrac{3}{2} = 1\\dfrac{1}{2}$$ +
Ответ: д) $1\\dfrac{1}{2}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "соответственные углы при двух параллельных прямых и секущей равны между собой;"], + ["б", "средняя линия трапеции параллельна основаниям;"], + ["в", "$\\cos 60^{\\circ} = \\dfrac{1}{2}$;"], + ["г", "если диагонали параллелограмма перпендикулярны, то это обязательно квадрат?"], + ], + sol: `
    +
  • а) Соответственные углы при ∥ прямых равны — верно
    • +
  • б) Средняя линия трапеции ∥ основаниям — верно
    • +
  • в) $\\cos 60^\\circ = \\frac{1}{2}$ — верно
    • +
  • г) Перпендикулярные диагонали → квадрат — НЕВЕРНО
    • +
+Если диагонали параллелограмма перпендикулярны, он является ромбом, но не обязательно квадратом. +
Ответ: г)
` + }, + { + text: `Определите наименьшее целое решение двойного неравенства + $-2 \\lt \\dfrac{2x - 1}{3} \\leq 1$.`, + sol: `Умножим все части на $3$: $-6 \\lt 2x-1 \\leq 3$. Прибавим $1$: $-5 \\lt 2x \\leq 4$. Разделим на $2$: $-2{,}5 \\lt x \\leq 2$. + + + + −3 + −2 + −1 + 0 + 1 + 2 + + + + −2,5 + +Наименьшее целое число, большее $-2{,}5$: это $-2$. +
Ответ: $-2$
` + }, + { + text: `В трапеции $ABCD$ с основаниями $AD$ и $BC$ точка $O$ — пересечение диагоналей, + $BC = 8$ см, $BO = 4$ см, $OD = 6$ см. Найдите среднюю линию трапеции.`, + sol: `Свойство диагоналей трапеции: точка пересечения диагоналей делит каждую диагональ в отношении, равном отношению оснований. +
Формула средней линии трапеции: $m = \\dfrac{a+b}{2}$ — полусумма оснований. +

+Шаг 1. По свойству диагоналей: +$$\\dfrac{BO}{OD} = \\dfrac{BC}{AD}$$ +Шаг 2. Подставляем $BO=4$, $OD=6$, $BC=8$: +$$\\dfrac{4}{6} = \\dfrac{8}{AD}$$ +$$AD = \\dfrac{8\\cdot 6}{4} = 12\\text{ см}$$ + + + + + + A + D + C + B + O + AD = 12 + BC = 8 + BO=4 + OD=6 + +Шаг 3. По формуле средней линии трапеции: +$$m = \\dfrac{BC+AD}{2} = \\dfrac{8+12}{2} = 10\\text{ см}$$ +
Ответ: $10$ см
` + }, + { + text: `Упростите выражение + $$\\dfrac{x + 2\\sqrt{xy} + y}{\\sqrt{x} + \\sqrt{y}} - \\dfrac{4x - y}{2\\sqrt{x} - \\sqrt{y}}.$$`, + sol: `Формула квадрата суммы: $(a+b)^2 = a^2+2ab+b^2$. +
Формула разности квадратов: $a^2-b^2 = (a-b)(a+b)$. +
Идея: представляем $x$ как $(\\sqrt{x})^2$, чтобы использовать формулы сокращённого умножения и сократить дроби. +

+Шаг 1. Преобразуем первую дробь. Числитель — полный квадрат: +$$x + 2\\sqrt{xy} + y = (\\sqrt{x})^2 + 2\\cdot\\sqrt{x}\\cdot\\sqrt{y} + (\\sqrt{y})^2 = (\\sqrt{x}+\\sqrt{y})^2$$ +Сокращаем на $(\\sqrt{x}+\\sqrt{y})$: +$$\\dfrac{(\\sqrt{x}+\\sqrt{y})^2}{\\sqrt{x}+\\sqrt{y}} = \\sqrt{x}+\\sqrt{y}$$ +Шаг 2. Преобразуем вторую дробь. Числитель — разность квадратов: +$$4x - y = (2\\sqrt{x})^2 - (\\sqrt{y})^2 = (2\\sqrt{x}-\\sqrt{y})(2\\sqrt{x}+\\sqrt{y})$$ +Сокращаем на $(2\\sqrt{x}-\\sqrt{y})$: +$$\\dfrac{(2\\sqrt{x}-\\sqrt{y})(2\\sqrt{x}+\\sqrt{y})}{2\\sqrt{x}-\\sqrt{y}} = 2\\sqrt{x}+\\sqrt{y}$$ +Шаг 3. Вычисляем разность: +$$(\\sqrt{x}+\\sqrt{y}) - (2\\sqrt{x}+\\sqrt{y}) = \\sqrt{x} + \\sqrt{y} - 2\\sqrt{x} - \\sqrt{y} = -\\sqrt{x}$$ +
Ответ: $-\\sqrt{x}$
` + }, + { + text: `График функции $y = g(x)$ получен из графика функции $f(x) = x^2$ + сдвигом на $2$ единицы вправо вдоль оси абсцисс и на $4$ единицы вниз + вдоль оси ординат. Найдите нули функции $y = g(x)$.`, + sol: `Правило сдвига графика функции: +
• сдвиг на $a$ единиц вправо: $f(x) \\to f(x-a)$; +
• сдвиг на $b$ единиц вниз: $f(x) \\to f(x)-b$. +
Нули функции: значения $x$, при которых $g(x)=0$. +

+Шаг 1. Сдвинем график $f(x)=x^2$ на $2$ единицы вправо. По правилу: +$$f_1(x) = (x-2)^2$$ +Шаг 2. Затем сдвинем график вниз на $4$ единицы: +$$g(x) = (x-2)^2 - 4$$ +Шаг 3. Найдём нули, решая уравнение $g(x)=0$: +$$(x-2)^2 - 4 = 0 \\implies (x-2)^2 = 4$$ +Шаг 4. Извлекаем квадратный корень: +$$x - 2 = \\pm 2$$ +Значит, $x = 4$ или $x = 0$. + + + + + xy + + + 0 + + 4 + + (2; −4) + + 2 + +
Ответ: $x = 0$ и $x = 4$
` + }, + { + text: `Решите уравнение $\\dfrac{7}{x^2 - x - 12} + \\dfrac{1}{x + 3} = -1$.`, + sol: `Решение дробно-рациональных уравнений: ищем ОДЗ, умножаем на общий знаменатель, проверяем корни. +
Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. Разложим знаменатель первой дроби. Сумма корней $1$, произведение $-12$. Подходят $4$ и $-3$: +$$x^2-x-12 = (x-4)(x+3)$$ +Шаг 2. ОДЗ: знаменатели не должны быть нулём: +$$x\\neq 4,\\quad x\\neq -3$$ +Шаг 3. Умножим обе части уравнения на общий знаменатель $(x-4)(x+3)$: +$$7 + (x-4) = -(x-4)(x+3)$$ +Шаг 4. Упрощаем: +$$x+3 = -(x^2-x-12) = -x^2+x+12$$ +$$x^2 - 9 = 0$$ +Шаг 5. По формуле разности квадратов: $(x-3)(x+3)=0 \\Rightarrow x=3$ или $x=-3$. +
Шаг 6. Проверка ОДЗ: $x=-3$ исключён. Проверяем $x=3$: +$$\\dfrac{7}{9-3-12}+\\dfrac{1}{6} = \\dfrac{7}{-6}+\\dfrac{1}{6} = -1 \\checkmark$$ +
Ответ: $x = 3$
` + }, + { + text: `Если двузначное число разделить на сумму его цифр, то в частном получится $4$ + и в остатке $3$. Если это же двузначное число разделить на произведение его цифр, + то в частном получится $3$ и в остатке $5$. Найдите это двузначное число.`, + sol: `Запись двузначного числа: $10a+b$, где $a$ — цифра десятков ($1\\leq a\\leq 9$), $b$ — цифра единиц ($0\\leq b\\leq 9$). +
Теорема о делении с остатком: если $N$ при делении на $d$ даёт частное $q$ и остаток $r$, то $N = d\\cdot q + r$, $0\\leq r\\lt d$. +
Формула корней квадратного уравнения: $x=\\dfrac{-b\\pm\\sqrt{D}}{2a}$, $D=b^2-4ac$. +

+Шаг 1. Обозначим за $10a+b$ искомое двузначное число. +
Шаг 2. Первое условие: при делении на $a+b$ получили частное $4$ и остаток $3$: +$$10a + b = 4(a+b) + 3$$ +$$10a + b = 4a + 4b + 3$$ +$$6a - 3b = 3$$ +$$b = 2a - 1 \\quad (*)$$ +Шаг 3. Второе условие: при делении на $ab$ получили частное $3$ и остаток $5$: +$$10a + b = 3ab + 5$$ +Шаг 4. Подставим $(*)$ во второе условие: +$$10a + (2a-1) = 3a(2a-1) + 5$$ +$$12a - 1 = 6a^2 - 3a + 5$$ +$$6a^2 - 15a + 6 = 0$$ +Делим на $3$: +$$2a^2 - 5a + 2 = 0$$ +Шаг 5. Решаем квадратное уравнение. Дискриминант: $D = 25 - 16 = 9 = 3^2$. +$$a = \\dfrac{5\\pm 3}{4} \\implies a = 2 \\text{ или } a = \\dfrac{1}{2}$$ +Цифра должна быть целой, значит $a = 2$. +
Шаг 6. Из $(*)$: $b = 2\\cdot 2 - 1 = 3$. Число $= 23$. +
Проверка: +
• сумма цифр $= 5$; $23:5 = 4$ (ост. $3$): $4\\cdot 5+3=23$ ✓; +
• произведение цифр $= 6$; $23:6 = 3$ (ост. $5$): $3\\cdot 6+5=23$ ✓. +
Ответ: $23$
` + }, + { + text: `Внутри параллелограмма $ABCD$ взята точка $K$, такая, что + $S_{BKC} = 12$ см², $S_{AKD} = 20$ см². + Найдите площадь параллелограмма $ABCD$.`, + figure: ` + + + + + + + + + A + B + C + D + K + 20 см² + 12 см² +`, + sol: `Свойство точки внутри параллелограмма: для любой точки $K$ внутри параллелограмма $ABCD$ сумма площадей двух треугольников, опирающихся на противоположные стороны, равна половине площади параллелограмма: +$$S_{\\triangle BKC} + S_{\\triangle AKD} = \\dfrac{S_{ABCD}}{2}$$ +
(Доказательство: $BC$ и $AD$ — противоположные стороны параллелограмма, $BC=AD=a$. Расстояния от $K$ до этих параллельных сторон $h_1$ и $h_2$ в сумме дают высоту параллелограмма $H$. Поэтому $S_{\\triangle BKC}+S_{\\triangle AKD}=\\tfrac{1}{2}a(h_1+h_2)=\\tfrac{1}{2}aH=\\tfrac{1}{2}S_{ABCD}$.) +

+Шаг 1. По свойству, доказанному выше: +$$\\dfrac{S_{ABCD}}{2} = S_{\\triangle BKC}+S_{\\triangle AKD}$$ +Шаг 2. Подставляем известные площади: +$$\\dfrac{S_{ABCD}}{2} = 12 + 20 = 32$$ +Шаг 3. Умножаем обе части на 2: +$$S_{ABCD} = 64\\text{ см}^2$$ +
Ответ: $64$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v09.js b/frontend/js/exam9/variants/v09.js new file mode 100644 index 0000000..9a4a44e --- /dev/null +++ b/frontend/js/exam9/variants/v09.js @@ -0,0 +1,251 @@ +VARIANTS[9] = { + label: "Вариант 9", + tasks: [ + { + text: `Определите неравенство, множество решений которого изображено на рисунке:`, + figure: ` + + + + + + −2 + x +`, + opts: [ + ["а", "$x < -2$"], ["б", "$x \\leq -2$"], ["в", "$x > -2$"], + ["г", "$x \\geq -2$"], ["д", "$x \\in \\mathbb{R}$"], + ], + sol: `На рисунке: луч идёт вправо от точки $-2$, точка закрашена (включена). +
Закрашенная точка → $\\geq$; луч вправо → $x \\geq -2$. +
Ответ: г) $x \\geq -2$
` + }, + { + text: `$15\\%$ от числа $30$ равны:`, + opts: [ + ["а", "$0{,}45$"], ["б", "$4{,}5$"], ["в", "$450$"], ["г", "$200$"], ["д", "$2$"], + ], + sol: `$$15\\% \\text{ от } 30 = \\frac{15}{100}\\cdot 30 = 0{,}15\\cdot 30 = 4{,}5$$ +
Ответ: б) $4{,}5$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "площадь квадрата со стороной $a$ равна $a^2$;"], + ["б", "сумма всех углов параллелограмма равна $360^{\\circ}$;"], + ["в", "синусом острого угла называется отношение прилежащего катета к гипотенузе;"], + ["г", "биссектриса равнобедренного треугольника, проведённая к основанию, является и медианой?"], + ], + sol: `
    +
  • а) Площадь квадрата $= a^2$ — верно
    • +
  • б) Сумма углов параллелограмма $= 360°$ — верно
    • +
  • в) НЕВЕРНО: $\\sin\\alpha = \\dfrac{\\text{противолежащий катет}}{\\text{гипотенуза}}$
    • +
  • г) Биссектриса к основанию равнобедренного ∆ = медиана — верно
    • +
+В утверждении в) описана косинус, а не синус: $\\cos\\alpha = \\dfrac{\\text{прилежащий катет}}{\\text{гипотенуза}}$. +
Ответ: в)
` + }, + { + text: `Приведите одночлен $4ab^2 \\cdot ab \\cdot b^4a \\cdot (-0{,}5)$ к стандартному виду.`, + sol: `Числовой коэффициент: $4\\cdot 1\\cdot 1\\cdot(-0{,}5) = -2$. +
Степени: $a^{1+1+1}=a^3$;  $b^{2+1+4}=b^7$. +$$4ab^2\\cdot ab\\cdot b^4a\\cdot(-0{,}5) = -2a^3b^7$$ +
Ответ: $-2a^3b^7$
` + }, + { + text: `Для квадратичной функции $y = -x^2 + 4x$ найдите значения аргумента, + при которых значение функции равно $3$.`, + sol: `Чтобы найти аргумент при заданном значении функции, надо приравнять формулу функции к этому значению и решить полученное уравнение относительно $x$. +
Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. По условию $y=3$. Приравняем формулу функции к $3$: +$$-x^2+4x = 3$$ +Шаг 2. Перенесём всё в одну часть, сменив знаки: +$$x^2-4x+3=0$$ +Шаг 3. По теореме Виета ищем корни: $x_1+x_2=4$, $x_1\\cdot x_2=3$. Подходят $1$ и $3$: +$$(x-1)(x-3)=0 \\implies x=1 \\text{ или } x=3$$ + + + + + xy + + + + + + 3 + + + 1 + + 3 + + (2; 4) + +
Ответ: $x = 1$ и $x = 3$
` + }, + { + text: `Треугольник $ABC$ — прямоугольный ($\\angle C = 90^{\\circ}$), $AC = 4$ см, + проекция катета $BC$ на гипотенузу равна $6$ см. + Найдите длину гипотенузы треугольника $ABC$.`, + sol: `Опустим высоту $CH$ из прямого угла $C$ на гипотенузу $AB$. + + + + + + + + + + + A + B + C + H + + AC=4 + + BH=6 + + AB=? + +Свойство высоты в прямоугольном треугольнике: +$$AC^2 = AH \\cdot AB \\quad \\text{и} \\quad BC^2 = BH \\cdot AB$$ +Из второго соотношения: $BH = \\dfrac{BC^2}{AB}$ — это и есть проекция $BC$ на гипотенузу $= 6$. +
Из первого: $AH = \\dfrac{AC^2}{AB} = \\dfrac{16}{AB}$. +
Так как $AH + BH = AB$: +$$\\dfrac{16}{AB} + 6 = AB$$ +$$AB^2 - 6\\cdot AB - 16 = 0$$ +$$(AB-8)(AB+2)=0 \\implies AB = 8\\text{ см}$$ +
Ответ: $8$ см
` + }, + { + text: `Решите систему уравнений + $$\\begin{cases} x - 4y = 2, \\\\[4pt] xy + 2y = 8. \\end{cases}$$`, + sol: `Метод подстановки для решения системы: из одного уравнения выражаем одну переменную через другую, подставляем в другое и решаем относительно одной переменной. +
Теорема Виета (обратная): $y^2+py+q=(y-y_1)(y-y_2)$, где $y_1+y_2=-p$, $y_1\\cdot y_2=q$. +

+Шаг 1. Из первого уравнения выразим $x$ через $y$: +$$x = 2 + 4y$$ +Шаг 2. Подставим $x$ во второе уравнение: +$$(2+4y)y + 2y = 8$$ +$$2y + 4y^2 + 2y = 8$$ +$$4y^2 + 4y - 8 = 0$$ +Шаг 3. Разделим обе части на $4$, чтобы упростить: +$$y^2 + y - 2 = 0$$ +Шаг 4. По теореме Виета: $y_1+y_2=-1$, $y_1\\cdot y_2=-2$. Подходят $-2$ и $1$: +$$(y+2)(y-1)=0 \\implies y=-2 \\text{ или } y=1$$ +Шаг 5. По формуле $x=2+4y$ находим $x$ для каждого $y$: +
$y=-2$:$x=2+4(-2)=-6$ → $(-6;\\,-2)$
$y=1$:$x=2+4(1)=6$ → $(6;\\,1)$
+Шаг 6. Проверка $(-6;-2)$: $-6-4(-2)=-6+8=2$ ✓; $(-6)(-2)+2(-2)=12-4=8$ ✓ +
Ответ: $(-6;\\,-2)$ и $(6;\\,1)$
` + }, + { + text: `Для перевозки $105$ т груза фирма рассматривала модели грузовых автомобилей МАЗ-4371СО. + Чтобы выполнить работы в срок, было решено использовать грузовой автомобиль + грузоподъёмностью на $2$ т больше. В результате для перевозки груза было сделано + на $6$ рейсов меньше, чем планировалось. + Найдите грузоподъёмность машины, на которой перевезли груз.`, + sol: `Пусть первоначальная грузоподъёмность $= p$ т. Число плановых рейсов $= \\dfrac{105}{p}$. +
Новая грузоподъёмность $= p+2$ т. Число фактических рейсов $= \\dfrac{105}{p+2}$. +
Условие — на $6$ рейсов меньше: +$$\\frac{105}{p} - \\frac{105}{p+2} = 6$$ +$$105\\cdot\\frac{(p+2)-p}{p(p+2)} = 6 \\implies \\frac{210}{p(p+2)}=6$$ +$$p(p+2) = 35 \\implies p^2+2p-35=0 \\implies (p+7)(p-5)=0$$ +$p=5$ (т.к. $p>0$). Грузоподъёмность использованной машины: $p+2 = 7$ т. +
Ответ: $7$ т
` + }, + { + text: `Определите количество целых решений неравенства + $$\\dfrac{(-x^2 - x + 6)\\,x^2}{x^2 - x - 2} \\geq 0.$$`, + sol: `Шаг 1 — разложим на множители. +$$-x^2-x+6 = -(x^2+x-6) = -(x+3)(x-2)$$ +$$x^2-x-2 = (x-2)(x+1)$$ +Выражение: $\\dfrac{-x^2(x+3)(x-2)}{(x-2)(x+1)}$. ОДЗ: $x\\neq 2$, $x\\neq -1$. +
Шаг 2 — сократим $(x-2)$ при $x\\neq 2$: +$$\\frac{-x^2(x+3)}{x+1} \\geq 0 \\iff \\frac{x^2(x+3)}{x+1} \\leq 0$$ +Шаг 3 — знаковый анализ. +$x^2 \\geq 0$ всегда. При $x=0$: выражение $=0$ ✓. При $x\\neq 0$: $x^2>0$, нужно $\\dfrac{x+3}{x+1}\\leq 0$. +
Критические точки: $x=-3$ (числитель $=0$), $x=-1$ (знаменатель $=0$, исключён). +
$\\dfrac{x+3}{x+1}\\leq 0$ выполняется при $-3\\leq x < -1$. + + + + −4 + −3 + −2 + −1 + 0 + 1 + + + + + + + +Решение: $-3\\leq x < -1$ или $x=0$. +
Целые числа: $x=-3,\\,-2,\\,0$ — итого 3. +
Ответ: $3$
` + }, + { + text: `В прямоугольную трапецию с основаниями $4$ и $8$ вписана окружность. + Найдите площадь трапеции.`, + sol: `Пусть $a=8$ (большее основание), $b=4$ (меньшее), $h$ — высота, $r$ — радиус окружности. + + + + + + + + + + + A + B + C + D + + a = 8 + + b = 4 + + h + + + r + + + +Шаг 1. Высота трапеции равна диаметру вписанной окружности: +$$h = 2r$$ +(окружность касается обоих оснований снизу и сверху, поэтому её диаметр = расстояние между ними). +

+Шаг 2. Найдём наклонную боковую сторону $AD$. +
Из свойства вписанной окружности в трапецию: сумма оснований = сумма боковых сторон: +$$AB + CD = BC + AD$$ +$$8 + 4 = h + AD$$ +$$AD = 12 - h = 12 - 2r$$ +
+Шаг 3. Применим теорему Пифагора к наклонной стороне $AD$. +
В прямоугольной трапеции $AD$ — гипотенуза прямоугольного треугольника с катетами $h$ и $(a-b)$: +$$AD^2 = h^2 + (a-b)^2$$ +Подставим $AD = 12-2r$ и $h=2r$, $a-b=8-4=4$: +$$(12-2r)^2 = (2r)^2 + 4^2$$ +$$144 - 48r + 4r^2 = 4r^2 + 16$$ +$$144 - 48r = 16$$ +$$48r = 128$$ +$$r = \\frac{128}{48} = \\frac{8}{3}\\text{ см}$$ +
+Шаг 4. Высота трапеции: +$$h = 2r = 2\\cdot\\frac{8}{3} = \\frac{16}{3}\\text{ см}$$ +
+Шаг 5. Площадь трапеции: +$$S = \\frac{a+b}{2}\\cdot h = \\frac{8+4}{2}\\cdot\\frac{16}{3} = \\frac{12}{2}\\cdot\\frac{16}{3} = 6\\cdot\\frac{16}{3} = \\frac{96}{3} = 32\\text{ см}^2$$ +
Ответ: $32$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v10.js b/frontend/js/exam9/variants/v10.js new file mode 100644 index 0000000..fe9a58b --- /dev/null +++ b/frontend/js/exam9/variants/v10.js @@ -0,0 +1,219 @@ +VARIANTS[10] = { + label: "Вариант 10", + tasks: [ + { + text: `Определите неравенство, множество решений которого изображено на рисунке:`, + figure: ` + + + + + + −4 + x +`, + opts: [ + ["а", "$x < -4$"], ["б", "$x \\leq -4$"], ["в", "$x > -4$"], + ["г", "$x \\geq -4$"], ["д", "$x \\in \\mathbb{R}$"], + ], + sol: `На рисунке: луч идёт вправо от точки $-4$, точка закрашена (включена). +
Закрашенная точка → $\\geq$; луч вправо → $x \\geq -4$. +
Ответ: г) $x \\geq -4$
` + }, + { + text: `$25\\%$ от числа $40$ равны:`, + opts: [ + ["а", "$1$"], ["б", "$10$"], ["в", "$1000$"], ["г", "$160$"], ["д", "$1{,}6$"], + ], + sol: `$$25\\% \\text{ от } 40 = \\frac{25}{100}\\cdot 40 = 0{,}25\\cdot 40 = 10$$ +
Ответ: б) $10$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "площадь прямоугольника со сторонами $a$ и $b$ равна $ab$;"], + ["б", "сумма всех углов трапеции равна $360^{\\circ}$;"], + ["в", "косинусом острого угла называется отношение противолежащего катета к гипотенузе;"], + ["г", "биссектриса равнобедренного треугольника, проведённая к основанию, является и высотой?"], + ], + sol: `
    +
  • а) Площадь прямоугольника $= ab$ — верно
    • +
  • б) Сумма углов трапеции $= 360°$ — верно
    • +
  • в) НЕВЕРНО: $\\cos\\alpha = \\dfrac{\\text{прилежащий катет}}{\\text{гипотенуза}}$, а не противолежащий
    • +
  • г) Биссектриса к основанию равнобедренного $\\triangle$ — также высота — верно
    • +
+В утверждении в) описан синус: $\\sin\\alpha = \\dfrac{\\text{противолежащий катет}}{\\text{гипотенуза}}$. +
Ответ: в)
` + }, + { + text: `Приведите одночлен $8nm^3 \\cdot nm \\cdot n^5 \\cdot (-0{,}5)$ к стандартному виду.`, + sol: `Числовой коэффициент: $8\\cdot 1\\cdot 1\\cdot(-0{,}5) = -4$. +
Степени: $n^{1+1+5}=n^7$;  $m^{3+1}=m^4$. +$$8nm^3\\cdot nm\\cdot n^5\\cdot(-0{,}5) = -4n^7m^4$$ +
Ответ: $-4n^7m^4$
` + }, + { + text: `Для квадратичной функции $y = -x^2 + 2x$ найдите значения аргумента, + при которых значение функции равно $-3$.`, + sol: `Чтобы найти аргумент по значению функции, приравниваем формулу функции к этому значению и решаем уравнение. +
Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. По условию $y=-3$. Приравняем формулу функции к $-3$: +$$-x^2+2x = -3$$ +Шаг 2. Перенесём всё в одну часть, поменяв знаки: +$$x^2-2x-3 = 0$$ +Шаг 3. По теореме Виета: $x_1+x_2=2$, $x_1\\cdot x_2=-3$. Подходят $3$ и $-1$: +$$(x-3)(x+1)=0 \\implies x=3 \\text{ или } x=-1$$ + + + + + xy + + + −3 + + −1 + + 3 + (1; 1) + +
Ответ: $x = -1$ и $x = 3$
` + }, + { + text: `Треугольник $ABC$ — прямоугольный ($\\angle C = 90^{\\circ}$), $BC = 6$ см, + проекция катета $AC$ на гипотенузу равна $5$ см. + Найдите длину гипотенузы треугольника $ABC$.`, + sol: `Свойство проекций катетов на гипотенузу (метрические соотношения в прямоугольном треугольнике): если $CH$ — высота, проведённая из прямого угла $C$ на гипотенузу $AB$, то +$$AC^2 = AH\\cdot AB,\\qquad BC^2 = BH\\cdot AB$$ +где $AH$ и $BH$ — проекции катетов $AC$ и $BC$ на гипотенузу. +

+Опустим высоту $CH$ из прямого угла $C$ на гипотенузу $AB$. + + + + + + A + B + C + H + AH=5 + BC=6 + +Шаг 1. Проекция катета $AC$ на гипотенузу — это $AH$, по условию $AH=5$ см. +
Шаг 2. Из соотношения $BC^2 = BH\\cdot AB$ выразим $BH$ через $AB$: +$$BH = \\dfrac{BC^2}{AB} = \\dfrac{36}{AB}$$ +Шаг 3. Так как $H$ лежит между $A$ и $B$, то $AH + BH = AB$: +$$5 + \\dfrac{36}{AB} = AB$$ +Шаг 4. Умножим обе части на $AB$ (так как $AB\\gt 0$): +$$AB^2 - 5\\cdot AB - 36 = 0$$ +Шаг 5. По теореме Виета: сумма корней $=5$, произведение $=-36$. Подходят $9$ и $-4$: +$$(AB-9)(AB+4) = 0$$ +Так как $AB\\gt 0$, выбираем $AB=9$ см. +
Ответ: $9$ см
` + }, + { + text: `Решите систему уравнений + $$\\begin{cases} x - 3y = 4, \\\\[4pt] xy - 7y = 6. \\end{cases}$$`, + sol: `Метод подстановки: выражаем одну переменную из одного уравнения и подставляем в другое. +
Теорема Виета (обратная): $y^2+py+q=(y-y_1)(y-y_2)$, где $y_1+y_2=-p$, $y_1\\cdot y_2=q$. +

+Шаг 1. Из первого уравнения выразим $x$ через $y$: +$$x = 4 + 3y$$ +Шаг 2. Подставим $x$ во второе уравнение: +$$(4+3y)y - 7y = 6$$ +$$4y + 3y^2 - 7y = 6$$ +$$3y^2 - 3y - 6 = 0$$ +Шаг 3. Разделим обе части на $3$: +$$y^2 - y - 2 = 0$$ +Шаг 4. По теореме Виета: $y_1+y_2=1$, $y_1\\cdot y_2=-2$. Подходят $2$ и $-1$: +$$(y-2)(y+1)=0 \\implies y=2 \\text{ или } y=-1$$ +Шаг 5. Для каждого $y$ находим $x$ по формуле $x=4+3y$: +
$y=2$:$x=4+6=10$ → $(10;\\,2)$
$y=-1$:$x=4-3=1$ → $(1;\\,-1)$
+Шаг 6. Проверка $(10;\\,2)$: $10-3\\cdot 2=4$ ✓; $10\\cdot 2-7\\cdot 2=20-14=6$ ✓ +
Ответ: $(10;\\,2)$ и $(1;\\,-1)$
` + }, + { + text: `Для перевозки $140$ т груза фирма рассматривала модели грузовых автомобилей МАЗ-5551. + После изучения условий аренды было решено использовать грузовой автомобиль + грузоподъёмностью на $3$ т меньше. В результате для перевозки груза + понадобилось сделать на $6$ рейсов больше, чем планировалось. + Найдите грузоподъёмность машины, на которой перевезли груз.`, + sol: `Пусть первоначальная грузоподъёмность $= p$ т. Число плановых рейсов $= \\dfrac{140}{p}$. +
Новая грузоподъёмность $= p-3$ т. Число фактических рейсов $= \\dfrac{140}{p-3}$. +
Условие — на $6$ рейсов больше: +$$\\frac{140}{p-3} - \\frac{140}{p} = 6$$ +$$140\\cdot\\frac{p-(p-3)}{p(p-3)} = 6 \\implies \\frac{420}{p(p-3)} = 6$$ +$$p(p-3) = 70 \\implies p^2-3p-70 = 0 \\implies (p-10)(p+7) = 0$$ +$p=10$ (т.к. $p \\gt 3$). Грузоподъёмность использованной машины: $p-3 = 7$ т. +
Ответ: $7$ т
` + }, + { + text: `Определите количество целых решений неравенства + $$\\dfrac{(-x^2 - 4x + 5)\\,x^2}{x^2 - 2x - 3} \\geq 0.$$`, + sol: `Шаг 1 — разложим на множители. +$$-x^2-4x+5 = -(x^2+4x-5) = -(x+5)(x-1)$$ +$$x^2-2x-3 = (x-3)(x+1)$$ +Выражение: $\\dfrac{-x^2(x+5)(x-1)}{(x-3)(x+1)}$. ОДЗ: $x\\neq 3$, $x\\neq -1$. +
Шаг 2 — перепишем условие. +$$\\frac{-x^2(x+5)(x-1)}{(x-3)(x+1)} \\geq 0 \\iff \\frac{x^2(x+5)(x-1)}{(x-3)(x+1)} \\leq 0$$ +Шаг 3 — знаковый анализ. +При $x=0$ или $x=1$: выражение $=0$ ✓. +При $x\\neq 0,\\,1$: $x^2 \\gt 0$, нужно $\\dfrac{(x+5)(x-1)}{(x-3)(x+1)} \\leq 0$. +
Критические точки: $x=-5$, $x=-1$ (искл.), $x=1$, $x=3$ (искл.). +
Методом интервалов (числитель меняет знак в $-5$ и $1$; знаменатель в $-1$ и $3$): + + + + −5 + −1 + 0 + 1 + 3 + + + + + + + + +Решение: $-5 \\leq x \\lt -1$, или $x=0$, или $1 \\leq x \\lt 3$. +
Целые числа: $x=-5,\\,-4,\\,-3,\\,-2$ и $x=0$ и $x=1,\\,2$ — итого 7. +
Ответ: $7$
` + }, + { + text: `В прямоугольную трапецию с основаниями $6$ и $12$ вписана окружность. + Найдите площадь трапеции.`, + sol: `Пусть $a=12$ (большее основание), $b=6$ (меньшее), $h$ — высота, $r$ — радиус вписанной окружности. + + + + + + A + B + C + D + a = 12 + b = 6 + h + + r + + +Шаг 1. Высота равна диаметру вписанной окружности: $h = 2r$. +
Шаг 2. Из свойства вписанной окружности в трапецию ($a+b = h+AD$): +$$12+6 = 2r+AD \\implies AD = 18-2r$$ +Шаг 3. Теорема Пифагора (наклонная боковая $AD$, разность оснований $a-b=6$): +$$(18-2r)^2 = (2r)^2+6^2$$ +$$324-72r+4r^2 = 4r^2+36$$ +$$72r = 288 \\implies r = 4\\text{ см}$$ +Шаг 4. Высота: $h = 2r = 8$ см. +
Шаг 5. Площадь: +$$S = \\dfrac{a+b}{2}\\cdot h = \\dfrac{12+6}{2}\\cdot 8 = 9\\cdot 8 = 72\\text{ см}^2$$ +
Ответ: $72$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v11.js b/frontend/js/exam9/variants/v11.js new file mode 100644 index 0000000..8ae0a3f --- /dev/null +++ b/frontend/js/exam9/variants/v11.js @@ -0,0 +1,213 @@ +VARIANTS[11] = { + label: "Вариант 11", + tasks: [ + { + text: `Определите, графиком какой из следующих функций является прямая:`, + opts: [ + ["а", "$y = \\dfrac{4}{x}$"], ["б", "$y = \\dfrac{x}{4}$"], ["в", "$y = 4x^2$"], + ["г", "$y = 4\\sqrt{x}$"], ["д", "$y = 4|x|$"], + ], + sol: `
    +
  • а) $y=4/x$ — гипербола
    • +
  • б) $y=x/4=(1/4)x$ — линейная функция вида $y=kx$, её график — прямая ✓
    • +
  • в) $y=4x^2$ — парабола
    • +
  • г) $y=4\\sqrt{x}$ — ветвь параболы (только $x\\geq 0$)
    • +
  • д) $y=4|x|$ — две полупрямые (V-образный график)
    • +
+
Ответ: б) $y=\\dfrac{x}{4}$
` + }, + { + text: `Выражение, тождественно равное выражению $(a^5)^{-1} \\cdot a^{-13}$, имеет вид:`, + opts: [ + ["а", "$a^{-9}$"], ["б", "$a^{-18}$"], ["в", "$a^8$"], + ["г", "$a^{18}$"], ["д", "$a^{17}$"], + ], + sol: `$$(a^5)^{-1}\\cdot a^{-13} = a^{-5}\\cdot a^{-13} = a^{-5+(-13)} = a^{-18}$$ +
Ответ: б) $a^{-18}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "любая сторона треугольника меньше суммы двух других его сторон;"], + ["б", "сумма углов треугольника равна $360^{\\circ}$;"], + ["в", "средняя линия трапеции равна полусумме оснований;"], + ["г", "на плоскости две прямые, параллельные третьей, параллельны между собой?"], + ], + sol: `
    +
  • а) Неравенство треугольника — верно
    • +
  • б) Сумма углов треугольника $= 360°$ — НЕВЕРНО
    • +
  • в) Средняя линия трапеции $=\\frac{a+b}{2}$ — верно
    • +
  • г) Транзитивность параллельности — верно
    • +
+Сумма углов любого треугольника равна $\\mathbf{180°}$, а не $360°$. +
Ответ: б)
` + }, + { + text: `Решите уравнение $0{,}64 - x^2 = 0$. + В ответ запишите наименьший корень уравнения.`, + sol: `$$x^2 = 0{,}64 = \\left(\\frac{8}{10}\\right)^2 \\implies x = \\pm 0{,}8$$ +Наименьший корень: $x = -0{,}8$. +
Ответ: $-0{,}8$
` + }, + { + text: `Угол между высотой прямоугольного треугольника, проведённой к гипотенузе, + и одним из катетов равен $60^{\\circ}$. Второй катет равен $16$ см. + Найдите радиус окружности, описанной около треугольника.`, + sol: `В прямоугольном $\\triangle ABC$ ($\\angle C=90°$) высота $CH$ опущена на гипотенузу $AB$. + + + + + + + 60° + A + B + C + H + AC=16 + 2R = AB = ? + +Свойство высоты в прямоугольном треугольнике: угол между высотой $CH$ и катетом $BC$ равен $\\angle A$. +$$\\angle BCH = \\angle A = 60°$$ +Значит $\\angle A=60°$, $\\angle B=30°$. +
«Второй катет» $AC$ — тот, что не связан напрямую с $60°$: +$$\\sin(\\angle B) = \\frac{AC}{AB} \\implies \\sin 30° = \\frac{16}{AB} \\implies \\frac{1}{2} = \\frac{16}{AB}$$ +$$AB = 32\\text{ см}$$ +Радиус описанной окружности прямоугольного треугольника: +$$R = \\frac{AB}{2} = \\frac{32}{2} = 16\\text{ см}$$ +
Ответ: $R = 16$ см
` + }, + { + text: `Найдите значение выражения + $\\dfrac{1}{1+\\sqrt{2}} + \\dfrac{1}{\\sqrt{2}+\\sqrt{3}} + \\dfrac{1}{\\sqrt{3}+2}$.`, + sol: `Метод рационализации знаменателя: чтобы избавиться от радикала в знаменателе, умножаем числитель и знаменатель на сопряжённое выражение — то же выражение с противоположным знаком. +
Формула разности квадратов: $(a-b)(a+b) = a^2-b^2$ — позволяет «сворачивать» произведение сопряжённых выражений. +
Идея: после рационализации возникает телескопическая сумма, в которой соседние слагаемые попарно сокращаются. +

+Шаг 1. Рационализируем первую дробь. Сопряжённое выражение для $1+\\sqrt{2}$ — это $\\sqrt{2}-1$. Умножим числитель и знаменатель на него: +$$\\dfrac{1}{1+\\sqrt{2}} = \\dfrac{\\sqrt{2}-1}{(1+\\sqrt{2})(\\sqrt{2}-1)} = \\dfrac{\\sqrt{2}-1}{(\\sqrt{2})^2-1^2} = \\dfrac{\\sqrt{2}-1}{2-1} = \\sqrt{2}-1$$ +Шаг 2. Рационализируем вторую дробь. Сопряжённое к $\\sqrt{2}+\\sqrt{3}$ — $\\sqrt{3}-\\sqrt{2}$: +$$\\dfrac{1}{\\sqrt{2}+\\sqrt{3}} = \\dfrac{\\sqrt{3}-\\sqrt{2}}{(\\sqrt{2}+\\sqrt{3})(\\sqrt{3}-\\sqrt{2})} = \\dfrac{\\sqrt{3}-\\sqrt{2}}{3-2} = \\sqrt{3}-\\sqrt{2}$$ +Шаг 3. Рационализируем третью дробь. Сопряжённое к $\\sqrt{3}+2$ — $2-\\sqrt{3}$: +$$\\dfrac{1}{\\sqrt{3}+2} = \\dfrac{2-\\sqrt{3}}{(\\sqrt{3}+2)(2-\\sqrt{3})} = \\dfrac{2-\\sqrt{3}}{4-3} = 2-\\sqrt{3}$$ +Шаг 4. Складываем полученные выражения. Заметим, что сумма телескопическая — соседние слагаемые сокращаются: +$$(\\sqrt{2}-1) + (\\sqrt{3}-\\sqrt{2}) + (2-\\sqrt{3})$$ +Группируем подобные: +$$= -1 + (\\sqrt{2}-\\sqrt{2}) + (\\sqrt{3}-\\sqrt{3}) + 2 = -1 + 0 + 0 + 2 = 1$$ +
Ответ: $1$
` + }, + { + text: `Определите количество целых решений системы неравенств + $$\\begin{cases} \\dfrac{x+2}{2} - 3 \\leq \\dfrac{x-3}{3}, \\\\[6pt] x^2 < 5x + 6. \\end{cases}$$`, + sol: `Решение системы неравенств: решаем каждое неравенство отдельно, затем берём пересечение множеств решений. +
Метод интервалов для квадратного неравенства: раскладываем квадратный трёхчлен на множители $(x-x_1)(x-x_2)$ и находим знаки на интервалах между корнями. +

+Шаг 1. Решаем первое неравенство. Умножим обе части на $6$ (общий знаменатель): +$$3(x+2) - 18 \\leq 2(x-3)$$ +$$3x + 6 - 18 \\leq 2x - 6$$ +$$3x - 12 \\leq 2x - 6$$ +$$x \\leq 6$$ +Шаг 2. Решаем второе неравенство. Перенесём всё влево: +$$x^2 - 5x - 6 \\lt 0$$ +Шаг 3. По теореме Виета: $x_1+x_2=5$, $x_1\\cdot x_2=-6$. Подходят $6$ и $-1$: +$$(x-6)(x+1) \\lt 0$$ +Произведение отрицательно, когда множители разных знаков, что выполнено при $-1\\lt x\\lt 6$. +
Шаг 4. Берём пересечение решений двух неравенств: $\\{x\\leq 6\\}\\cap\\{-1\\lt x\\lt 6\\} = \\{-1\\lt x\\lt 6\\}$. +
Шаг 5. Считаем целые числа из промежутка $(-1;6)$: $0,1,2,3,4,5$ — всего 6 чисел. +
Ответ: $6$
` + }, + { + text: `Найдите все значения переменной, при которых разность дробей + $\\dfrac{x}{x+1}$ и $\\dfrac{1}{x}$ равна дроби $\\dfrac{1}{x^2+x}$.`, + sol: `Решение дробно-рациональных уравнений: 1) найти ОДЗ (знаменатели $\\neq 0$); 2) привести к общему знаменателю или умножить обе части на него; 3) решить полученное уравнение; 4) проверить, входят ли корни в ОДЗ. +
Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. Запишем условие в виде уравнения: +$$\\dfrac{x}{x+1} - \\dfrac{1}{x} = \\dfrac{1}{x^2+x}$$ +Шаг 2. Разложим знаменатель правой части: $x^2+x = x(x+1)$. Это и есть общий знаменатель всех трёх дробей. +
ОДЗ: знаменатели не равны нулю, поэтому $x\\neq 0$ и $x\\neq -1$. +
Шаг 3. Умножим обе части уравнения на $x(x+1)$: +$$\\dfrac{x}{x+1}\\cdot x(x+1) - \\dfrac{1}{x}\\cdot x(x+1) = \\dfrac{1}{x(x+1)}\\cdot x(x+1)$$ +$$x\\cdot x - (x+1)\\cdot 1 = 1$$ +$$x^2 - (x+1) = 1$$ +Шаг 4. Раскрываем скобки и приводим к стандартному виду: +$$x^2 - x - 1 = 1$$ +$$x^2 - x - 2 = 0$$ +Шаг 5. По теореме Виета: $x_1+x_2=1$, $x_1\\cdot x_2=-2$. Подходят $-1$ и $2$: +$$(x+1)(x-2) = 0 \\implies x = -1 \\text{ или } x = 2$$ +Шаг 6. Проверяем ОДЗ: $x=-1$ не входит в ОДЗ (отбрасываем). Остаётся $x=2$. +
Проверка подстановкой $x=2$: +$$\\dfrac{2}{3} - \\dfrac{1}{2} = \\dfrac{4-3}{6} = \\dfrac{1}{6};\\quad \\dfrac{1}{4+2} = \\dfrac{1}{6} \\checkmark$$ +
Ответ: $x = 2$
` + }, + { + text: `Число $a$ равно $70\\%$ от числа $b$, число $c$ на $42$ больше числа $b$. + Найдите значение выражения $a + b + c$, + если известно, что число $a$ равно $40\\%$ от числа $c$.`, + sol: `Перевод процентов в дроби: $p\\%$ от числа $N$ — это $\\dfrac{p}{100}\\cdot N$. +
Метод составления уравнения: переводим условия с процентами в равенства, получаем систему и решаем её. +

+Шаг 1. Запишем каждое условие в виде уравнения. +
• «$a$ равно $70\\%$ от $b$»: +$$a = 0{,}7b \\quad (1)$$ +
• «$c$ на $42$ больше $b$»: +$$c = b + 42 \\quad (2)$$ +
• «$a$ равно $40\\%$ от $c$»: +$$a = 0{,}4c \\quad (3)$$ +Шаг 2. Подставляем $(1)$ и $(2)$ в $(3)$, чтобы получить уравнение с одной неизвестной $b$: +$$0{,}7b = 0{,}4(b+42)$$ +Шаг 3. Раскрываем скобки: +$$0{,}7b = 0{,}4b + 16{,}8$$ +$$0{,}3b = 16{,}8$$ +$$b = 56$$ +Шаг 4. Находим $a$ из $(1)$ и $c$ из $(2)$: +$$a = 0{,}7\\cdot 56 = 39{,}2$$ +$$c = 56 + 42 = 98$$ +Проверка $(3)$: $0{,}4\\cdot 98 = 39{,}2 = a$ ✓ +
Шаг 5. Вычисляем сумму: +$$a + b + c = 39{,}2 + 56 + 98 = 193{,}2$$ +
Ответ: $193{,}2$
` + }, + { + text: `Дан параллелограмм $ABCD$. На стороне $BC$ взята точка $K$, такая, что + $AK$ — биссектриса угла $A$, а $DK$ — биссектриса угла $D$ параллелограмма. + Найдите площадь параллелограмма, если $AK = 8$ см, $DK = 6$ см.`, + figure: ` + + + + + A + B + C + D + K + AK = 8 см + DK = 6 см +`, + sol: `Шаг 1 — угол при K в треугольнике AKD. +
В параллелограмме $\\angle A + \\angle D = 180°$. Делим на 2: +$$\\angle KAD + \\angle KDA = \\frac{\\angle A}{2}+\\frac{\\angle D}{2} = 90°$$ +В $\\triangle AKD$ сумма углов $= 180°$: +$$\\angle AKD = 180° - (\\angle KAD + \\angle KDA) = 180° - 90° = \\mathbf{90°}$$ +Треугольник $AKD$ — прямоугольный! +

+Шаг 2 — длина AD. +$$AD = \\sqrt{AK^2 + DK^2} = \\sqrt{64+36} = \\sqrt{100} = 10\\text{ см}$$ +Шаг 3 — длина AB. +
В $\\triangle ABK$: $\\angle BAK = \\angle A/2$, $\\angle ABK = 180°-\\angle A$, поэтому $\\angle AKB = \\angle A/2$. +
Треугольник $ABK$ — равнобедренный: $BK = AB$. +
Аналогично в $\\triangle DKC$: $KC = DC = AB$. +$$BC = BK + KC = AB + AB = 2\\cdot AB$$ +Так как $BC = AD = 10$: $\\quad AB = 5$ см. +

+Шаг 4 — площадь. +
Из $\\triangle AKD$: $\\sin(\\angle KAD) = \\dfrac{DK}{AD} = \\dfrac{6}{10} = \\dfrac{3}{5}$, $\\cos(\\angle KAD) = \\dfrac{4}{5}$. +$$\\sin(\\angle A) = \\sin(2\\angle KAD) = 2\\cdot\\frac{3}{5}\\cdot\\frac{4}{5} = \\frac{24}{25}$$ +$$S = AB\\cdot AD\\cdot\\sin(\\angle A) = 5\\cdot 10\\cdot\\frac{24}{25} = \\frac{1200}{25} = 48\\text{ см}^2$$ +Проверка через части: $S_{ABK}+S_{AKD}+S_{DKC} = 12+24+12 = 48$ ✓ +
Ответ: $48$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v12.js b/frontend/js/exam9/variants/v12.js new file mode 100644 index 0000000..32c7e35 --- /dev/null +++ b/frontend/js/exam9/variants/v12.js @@ -0,0 +1,212 @@ +VARIANTS[12] = { + label: "Вариант 12", + tasks: [ + { + text: `Определите, графиком какой из следующих функций является прямая:`, + opts: [ + ["а", "$y = -\\dfrac{1}{5x}$"], ["б", "$y = \\dfrac{1}{5}x^2$"], ["в", "$y = -\\dfrac{1}{5}x$"], + ["г", "$y = \\dfrac{\\sqrt{x}}{5}$"], ["д", "$y = 5|x|$"], + ], + sol: `
    +
  • а) $y=-\\dfrac{1}{5x}$ — гипербола
    • +
  • б) $y=\\dfrac{1}{5}x^2$ — парабола
    • +
  • в) $y=-\\dfrac{1}{5}x = \\left(-\\dfrac{1}{5}\\right)x$ — линейная функция вида $y=kx$, её график — прямая ✓
    • +
  • г) $y=\\dfrac{\\sqrt{x}}{5}$ — ветвь параболы (только $x\\geq 0$)
    • +
  • д) $y=5|x|$ — V-образный график (две полупрямые)
    • +
+
Ответ: в) $y=-\\dfrac{1}{5}x$
` + }, + { + text: `Выражение, тождественно равное выражению $(b^{-2})^{-4} \\cdot b^3$, имеет вид:`, + opts: [ + ["а", "$b^{-3}$"], ["б", "$b^{-5}$"], ["в", "$b^{11}$"], + ["г", "$b^5$"], ["д", "$b^{-9}$"], + ], + sol: `$$(b^{-2})^{-4}\\cdot b^3 = b^{(-2)\\cdot(-4)}\\cdot b^3 = b^8\\cdot b^3 = b^{8+3} = b^{11}$$ +
Ответ: в) $b^{11}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "гипотенуза прямоугольного треугольника всегда больше катета;"], + ["б", "сумма углов четырёхугольника равна $180^{\\circ}$;"], + ["в", "средняя линия треугольника равна половине основания;"], + ["г", "на плоскости две прямые, перпендикулярные третьей, параллельны между собой?"], + ], + sol: `
    +
  • а) Гипотенуза — наибольшая сторона прямоугольного треугольника — верно
    • +
  • б) Сумма углов четырёхугольника $= 180°$ — НЕВЕРНО
    • +
  • в) Средняя линия треугольника $= \\dfrac{1}{2}$ основания — верно
    • +
  • г) Транзитивность перпендикулярности на плоскости — верно
    • +
+Сумма углов любого четырёхугольника равна $\\mathbf{360°}$, а не $180°$. +
Ответ: б)
` + }, + { + text: `Решите уравнение $0{,}49 - x^2 = 0$. + В ответ запишите наибольший корень уравнения.`, + sol: `$$x^2 = 0{,}49 = \\left(\\frac{7}{10}\\right)^2 \\implies x = \\pm 0{,}7$$ +Наибольший корень: $x = 0{,}7$. +
Ответ: $0{,}7$
` + }, + { + text: `Угол между высотой прямоугольного треугольника, проведённой к гипотенузе, + и катетом, длина которого $12$ см, равен $30^{\\circ}$. + Найдите радиус окружности, описанной около треугольника.`, + sol: `В прямоугольном $\\triangle ABC$ ($\\angle C=90°$) высота $CH$ опущена на гипотенузу $AB$. + + + + + + + 30° + A + B + C + H + BC=12 + 2R = AB = ? + +Свойство высоты в прямоугольном треугольнике: угол между высотой $CH$ и катетом $BC$ равен $\\angle A$. +$$\\angle BCH = \\angle A = 30°$$ +Значит $\\angle A=30°$, $\\angle B=60°$. +
Катет $BC=12$ см, $\\angle A=30°$: +$$\\sin(\\angle A) = \\dfrac{BC}{AB} \\implies \\sin 30° = \\dfrac{12}{AB} \\implies \\dfrac{1}{2} = \\dfrac{12}{AB}$$ +$$AB = 24\\text{ см}$$ +Радиус описанной окружности прямоугольного треугольника: +$$R = \\dfrac{AB}{2} = \\dfrac{24}{2} = 12\\text{ см}$$ +
Ответ: $R = 12$ см
` + }, + { + text: `Найдите значение выражения + $\\dfrac{1}{2-\\sqrt{3}} + \\dfrac{1}{\\sqrt{2}-\\sqrt{3}} + \\dfrac{1}{\\sqrt{2}-1}$.`, + sol: `Метод рационализации знаменателя: чтобы избавиться от радикала в знаменателе, умножаем числитель и знаменатель на сопряжённое выражение. +
Формула разности квадратов: $(a-b)(a+b) = a^2-b^2$ — позволяет «сворачивать» произведение сопряжённых. +

+Шаг 1. Рационализируем первую дробь. Сопряжённое к $2-\\sqrt{3}$ — это $2+\\sqrt{3}$: +$$\\dfrac{1}{2-\\sqrt{3}} = \\dfrac{2+\\sqrt{3}}{(2-\\sqrt{3})(2+\\sqrt{3})} = \\dfrac{2+\\sqrt{3}}{4-3} = 2+\\sqrt{3}$$ +Шаг 2. Рационализируем вторую дробь. Сопряжённое к $\\sqrt{2}-\\sqrt{3}$ — это $\\sqrt{2}+\\sqrt{3}$: +$$\\dfrac{1}{\\sqrt{2}-\\sqrt{3}} = \\dfrac{\\sqrt{2}+\\sqrt{3}}{(\\sqrt{2})^2-(\\sqrt{3})^2} = \\dfrac{\\sqrt{2}+\\sqrt{3}}{2-3} = -(\\sqrt{2}+\\sqrt{3})$$ +Шаг 3. Рационализируем третью дробь. Сопряжённое к $\\sqrt{2}-1$ — это $\\sqrt{2}+1$: +$$\\dfrac{1}{\\sqrt{2}-1} = \\dfrac{\\sqrt{2}+1}{(\\sqrt{2}-1)(\\sqrt{2}+1)} = \\dfrac{\\sqrt{2}+1}{2-1} = \\sqrt{2}+1$$ +Шаг 4. Складываем полученные выражения и группируем подобные: +$$(2+\\sqrt{3}) + \\bigl(-(\\sqrt{2}+\\sqrt{3})\\bigr) + (\\sqrt{2}+1)$$ +$$= 2 + \\sqrt{3} - \\sqrt{2} - \\sqrt{3} + \\sqrt{2} + 1$$ +$$= 2 + 1 + (\\sqrt{3}-\\sqrt{3}) + (-\\sqrt{2}+\\sqrt{2}) = 3$$ +
Ответ: $3$
` + }, + { + text: `Определите количество целых решений системы неравенств + $$\\begin{cases} x^2 \\leq 6 - x, \\\\[6pt] \\dfrac{x+3}{2} - 1 > \\dfrac{x-4}{7}. \\end{cases}$$`, + sol: `Решение системы неравенств: решаем каждое неравенство отдельно, затем берём пересечение множеств решений. +
Метод интервалов для квадратного неравенства: раскладываем квадратный трёхчлен и определяем знаки. +

+Шаг 1. Решаем первое неравенство. Перенесём $6-x$ влево: +$$x^2 + x - 6 \\leq 0$$ +По теореме Виета корни — $-3$ и $2$ (так как $x_1+x_2=-1$, $x_1\\cdot x_2=-6$). Раскладываем: +$$(x+3)(x-2) \\leq 0$$ +Произведение неположительно при $-3\\leq x\\leq 2$. +
Шаг 2. Решаем второе неравенство. Умножим обе части на $14$ (общий знаменатель): +$$7(x+3) - 14 \\gt 2(x-4)$$ +$$7x + 21 - 14 \\gt 2x - 8$$ +$$7x + 7 \\gt 2x - 8$$ +$$5x \\gt -15$$ +$$x \\gt -3$$ +Шаг 3. Берём пересечение: $\\{-3\\leq x\\leq 2\\}\\cap\\{x\\gt -3\\} = \\{-3\\lt x\\leq 2\\}$. +
Шаг 4. Целые числа из $(-3;2]$: $-2,-1,0,1,2$ — всего 5 чисел. +
Ответ: $5$
` + }, + { + text: `Найдите все значения переменной, при которых разность дробей + $\\dfrac{x}{x-2}$ и $\\dfrac{1}{x}$ равна дроби $\\dfrac{4}{x^2-2x}$.`, + sol: `Решение дробно-рациональных уравнений: 1) находим ОДЗ; 2) умножаем обе части на общий знаменатель; 3) решаем полученное уравнение; 4) проверяем корни на принадлежность ОДЗ. +
Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. Запишем уравнение: +$$\\dfrac{x}{x-2} - \\dfrac{1}{x} = \\dfrac{4}{x^2-2x}$$ +Шаг 2. Разложим знаменатель правой части: $x^2-2x = x(x-2)$ — это и есть общий знаменатель. +
ОДЗ: $x\\neq 0$ и $x\\neq 2$. +
Шаг 3. Умножим обе части на $x(x-2)$: +$$\\dfrac{x}{x-2}\\cdot x(x-2) - \\dfrac{1}{x}\\cdot x(x-2) = \\dfrac{4}{x(x-2)}\\cdot x(x-2)$$ +$$x\\cdot x - (x-2) = 4$$ +$$x^2 - x + 2 = 4$$ +$$x^2 - x - 2 = 0$$ +Шаг 4. По теореме Виета: $x_1+x_2=1$, $x_1\\cdot x_2=-2$. Подходят $2$ и $-1$: +$$(x-2)(x+1) = 0 \\implies x=2 \\text{ или } x=-1$$ +Шаг 5. Проверяем ОДЗ: $x=2$ не входит (отбрасываем). Остаётся $x=-1$. +
Проверка $x=-1$: +$$\\dfrac{-1}{-3} - \\dfrac{1}{-1} = \\dfrac{1}{3} + 1 = \\dfrac{4}{3};\\quad \\dfrac{4}{1+2} = \\dfrac{4}{3} \\checkmark$$ +
Ответ: $x = -1$
` + }, + { + text: `Число $a$ равно $80\\%$ от числа $b$. Число $c$ равно $140\\%$ от числа $b$. + Найдите значение выражения $a + b + c$, + если известно, что число $c$ на $72$ больше числа $a$.`, + sol: `Перевод процентов в дроби: $p\\%$ от числа $N$ — это $\\dfrac{p}{100}\\cdot N$. +
Метод составления уравнения: переводим условия с процентами в равенства и решаем. +

+Шаг 1. Запишем условия в виде уравнений. +
• «$a$ равно $80\\%$ от $b$»: +$$a = 0{,}8b \\quad (1)$$ +
• «$c$ равно $140\\%$ от $b$»: +$$c = 1{,}4b \\quad (2)$$ +
• «$c$ на $72$ больше $a$»: +$$c - a = 72 \\quad (3)$$ +Шаг 2. Подставляем $(1)$ и $(2)$ в $(3)$: +$$1{,}4b - 0{,}8b = 72$$ +$$0{,}6b = 72$$ +$$b = 120$$ +Шаг 3. Находим $a$ из $(1)$ и $c$ из $(2)$: +$$a = 0{,}8\\cdot 120 = 96$$ +$$c = 1{,}4\\cdot 120 = 168$$ +Проверка $(3)$: $c - a = 168 - 96 = 72$ ✓ +
Шаг 4. Вычисляем сумму: +$$a + b + c = 96 + 120 + 168 = 384$$ +
Ответ: $384$
` + }, + { + text: `Дан параллелограмм $ABCD$. На стороне $AD$ взята точка $M$, такая, что + $BM$ — биссектриса угла $B$, а $CM$ — биссектриса угла $C$ параллелограмма. + Найдите площадь параллелограмма, если $BM = 12$ см, $CM = 16$ см.`, + figure: ` + + + + + A + B + C + D + M + BM = 12 см + CM = 16 см +`, + sol: `Свойство параллелограмма: сумма углов, прилежащих к одной стороне, равна $180°$ (соседние углы дополнительны). +
Свойство биссектрис: биссектриса делит угол пополам. +
Теорема Пифагора: в прямоугольном треугольнике $c^2 = a^2+b^2$. +
Формула площади параллелограмма: $S = a\\cdot b\\cdot\\sin\\alpha$, где $\\alpha$ — угол между сторонами $a$ и $b$. +
Формула синуса двойного угла: $\\sin 2\\alpha = 2\\sin\\alpha\\cos\\alpha$. +

+Шаг 1 — найдём угол BMC. В параллелограмме $\\angle B + \\angle C = 180°$ (соседние углы). Делим на 2: +$$\\angle MBC + \\angle MCB = \\dfrac{\\angle B}{2} + \\dfrac{\\angle C}{2} = 90°$$ +В $\\triangle BMC$ сумма углов $180°$, значит: +$$\\angle BMC = 180° - 90° = 90°$$ +Треугольник $BMC$ — прямоугольный! +
Шаг 2 — длина BC. По теореме Пифагора: +$$BC = \\sqrt{BM^2 + CM^2} = \\sqrt{144+256} = \\sqrt{400} = 20\\text{ см}$$ +Шаг 3 — длина AB. В $\\triangle ABM$: углы $\\angle BAM = \\angle A/2$ и $\\angle AMB = \\angle MAD = \\angle A/2$ (накрест лежащие при параллельных $BC$ и $AD$). Значит, $\\triangle ABM$ — равнобедренный, и $AM = AB$. +
Аналогично в $\\triangle DCM$: $DM = DC = AB$. +
Так как $AD = AM + MD$: +$$AD = AB + AB = 2\\cdot AB$$ +$AD = BC = 20$, значит $AB = 10$ см. +
Шаг 4 — синус угла B. Из прямоугольного $\\triangle BMC$: +$$\\sin(\\angle MBC) = \\dfrac{CM}{BC} = \\dfrac{16}{20} = \\dfrac{4}{5},\\quad \\cos(\\angle MBC) = \\dfrac{BM}{BC} = \\dfrac{12}{20} = \\dfrac{3}{5}$$ +Поскольку $\\angle B = 2\\angle MBC$, применяем формулу двойного угла: +$$\\sin(\\angle B) = 2\\sin(\\angle MBC)\\cos(\\angle MBC) = 2\\cdot\\dfrac{4}{5}\\cdot\\dfrac{3}{5} = \\dfrac{24}{25}$$ +Шаг 5 — площадь. +$$S = AB\\cdot BC\\cdot\\sin(\\angle B) = 10\\cdot 20\\cdot\\dfrac{24}{25} = \\dfrac{4800}{25} = 192\\text{ см}^2$$ +
Ответ: $192$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v13.js b/frontend/js/exam9/variants/v13.js new file mode 100644 index 0000000..d72eb73 --- /dev/null +++ b/frontend/js/exam9/variants/v13.js @@ -0,0 +1,234 @@ +VARIANTS[13] = { + label: "Вариант 13", + tasks: [ + { + text: `Определите, какая из следующих пар чисел является вершиной параболы + $y = -(x-3)^2 + 1$:`, + opts: [ + ["а", "$(3;\\ 1)$"], ["б", "$(3;\\ {-1})$"], ["в", "$(-3;\\ 1)$"], + ["г", "$(-1;\\ 3)$"], ["д", "$(1;\\ {-3})$"], + ], + sol: `Парабола $y = a(x-h)^2 + k$ имеет вершину $(h;\\,k)$. +
Здесь $a=-1$, $h=3$, $k=1$ — вершина $(3;\\,1)$. + + + + + xy + + + (3; 1) + 1 + + 3 + +
Ответ: а) $(3;\\,1)$
` + }, + { + text: `Определите, какие из чисел НЕ являются решением двойного неравенства + $-6 < 2x < 8$:`, + opts: [ + ["а", "$-2$"], ["б", "$0$"], ["в", "$5$"], ["г", "$-3$"], ["д", "$1$"], + ], + sol: `Разделим на $2$: $-3 < x < 4$. +
Проверяем каждое число: +
    +
  • а) $-2$: $-3<-2<4$ ✓ — решение
    • +
  • б) $0$: ✓ — решение
    • +
  • в) $5$: $5>4$ ✗ — НЕ решение
    • +
  • г) $-3$: $-3$ не строго больше $-3$ ✗ — НЕ решение
    • +
  • д) $1$: ✓ — решение
    • +
+
Ответ: в) и г)
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "прямые, содержащие высоты треугольника, пересекаются в одной точке;"], + ["б", "периметр квадрата со стороной $a$ равен $4a$;"], + ["в", "углы при основании равнобедренной трапеции равны между собой;"], + ["г", "у любого параллелограмма все углы равны между собой?"], + ], + sol: `
    +
  • а) Высоты пересекаются в ортоцентре — верно
    • +
  • б) Периметр квадрата $= 4a$ — верно
    • +
  • в) Углы при основании равнобедренной трапеции равны — верно
    • +
  • г) Все углы параллелограмма равны — НЕВЕРНО
    • +
+В параллелограмме противоположные углы равны, а соседние в сумме дают $180°$. Равны все только у прямоугольника. +
Ответ: г)
` + }, + { + text: `Найдите значение выражения $6x - 12$, + если $\\dfrac{3x^2 - 5x}{x} = 0$.`, + sol: `При $x\\neq 0$: +$$\\frac{3x^2-5x}{x} = \\frac{x(3x-5)}{x} = 3x-5 = 0 \\implies x = \\frac{5}{3}$$ +$$6x-12 = 6\\cdot\\frac{5}{3}-12 = 10-12 = -2$$ +
Ответ: $-2$
` + }, + { + text: `Периметр ромба $ABCD$ равен $48$ см. Угол между стороной $AD$ и диагональю $AC$ + равен $30^{\\circ}$. Найдите диагональ $BD$ ромба.`, + sol: `Свойства ромба: все четыре стороны равны; диагонали перпендикулярны, делятся точкой пересечения пополам и делят углы ромба пополам (являются биссектрисами). +
Признак равностороннего треугольника: равнобедренный треугольник с углом $60°$ является равносторонним. +

+Шаг 1. Так как все стороны ромба равны, найдём сторону через периметр: +$$a = \\dfrac{P}{4} = \\dfrac{48}{4} = 12\\text{ см}$$ + + + + + + + + + + + + + 30° + + + + 60° + + A + B + C + D + + 12 + AC + BD = ? + +Шаг 2. Диагональ $AC$ — биссектриса угла $A$ ромба, поэтому $\\angle DAC = \\dfrac{\\angle DAB}{2}$. По условию $\\angle DAC = 30°$, значит: +$$\\angle DAB = 2\\cdot 30° = 60°$$ +Шаг 3. Рассмотрим треугольник $ABD$: +
    +
  • $AB = AD = 12$ (стороны ромба равны),
    • +
  • $\\angle DAB = 60°$.
    • +
+Это равнобедренный треугольник с углом $60°$ — значит, он равносторонний, и все стороны равны: +$$BD = AB = 12\\text{ см}$$ +Проверка по теореме косинусов: $BD^2 = 12^2+12^2-2\\cdot 12\\cdot 12\\cdot\\cos 60° = 144$ ✓ +
Ответ: $BD = 12$ см
` + }, + { + text: `Сократите дробь $\\dfrac{4n^2 - 9m^2}{9m^2 + 4n^2 + 12mn}$.`, + sol: `Формула разности квадратов: $a^2-b^2 = (a-b)(a+b)$. +
Формула квадрата суммы: $(a+b)^2 = a^2+2ab+b^2$. +

+Шаг 1. Разложим числитель по формуле разности квадратов ($a=2n$, $b=3m$): +$$4n^2 - 9m^2 = (2n)^2 - (3m)^2 = (2n-3m)(2n+3m)$$ +Шаг 2. Разложим знаменатель. Замечаем, что это квадрат суммы: +$$9m^2 + 4n^2 + 12mn = (3m)^2 + 2\\cdot 3m\\cdot 2n + (2n)^2 = (3m+2n)^2$$ +Так как $3m+2n = 2n+3m$, можно записать: +$$(3m+2n)^2 = (2n+3m)^2$$ +Шаг 3. Сокращаем общий множитель $(2n+3m)$: +$$\\dfrac{(2n-3m)(2n+3m)}{(2n+3m)^2} = \\dfrac{2n-3m}{2n+3m}$$ +
Ответ: $\\dfrac{2n-3m}{2n+3m}$
` + }, + { + text: `Решите систему уравнений + $$\\begin{cases} x - 4y = 2, \\\\[4pt] xy + 2y = 8 \\end{cases}$$ + и найдите значение выражения $x_1 \\cdot y_1 + x_2 \\cdot y_2$, + где $(x_1;\\, y_1)$, $(x_2;\\, y_2)$ — решения системы.`, + sol: `Метод подстановки: выражаем одну переменную через другую и подставляем. +
Теорема Виета (обратная): $y^2+py+q=(y-y_1)(y-y_2)$, где $y_1+y_2=-p$, $y_1\\cdot y_2=q$. +

+Шаг 1. Из первого уравнения выразим $x$: +$$x = 2 + 4y$$ +Шаг 2. Подставим во второе уравнение: +$$(2+4y)y + 2y = 8$$ +$$2y + 4y^2 + 2y = 8$$ +$$4y^2 + 4y - 8 = 0$$ +Шаг 3. Разделим на $4$: +$$y^2 + y - 2 = 0$$ +Шаг 4. По теореме Виета: $y_1+y_2=-1$, $y_1\\cdot y_2=-2$. Подходят $-2$ и $1$: +$$(y+2)(y-1)=0$$ +Шаг 5. Для каждого $y$ находим $x = 2+4y$: +
$y_1=-2$:$x_1=2-8=-6$
$y_2=1$:$x_2=2+4=6$
+Шаг 6. Вычисляем требуемое выражение: +$$x_1y_1+x_2y_2 = (-6)(-2)+(6)(1) = 12+6 = 18$$ +
Ответ: $18$
` + }, + { + text: `Найдите значение выражения $\\sqrt{8 - 2\\sqrt{7}} + \\sqrt{32 - 10\\sqrt{7}}$. + В ответ запишите число, обратное полученному.`, + sol: `Формула квадрата разности: $(a-b)^2 = a^2 - 2ab + b^2$. +
Формула выноса корня из квадрата: $\\sqrt{a^2}=|a|$. При $a\\geq 0$ модуль раскрывается как $|a|=a$, при $a\\lt 0$ — как $|a|=-a$. +
Идея: подкоренное выражение вида $A - 2\\sqrt{B}$ часто можно представить как квадрат разности $(\\sqrt{m}-\\sqrt{n})^2$, где $m+n=A$ и $mn=B$. +

+Шаг 1. Преобразуем первое подкоренное выражение: $8 - 2\\sqrt{7}$. +
Ищем представление в виде $(\\sqrt{m}-\\sqrt{n})^2 = m + n - 2\\sqrt{mn}$. Нужно $m+n=8$ и $mn=7$. Подходят $m=7$, $n=1$: +$$8 - 2\\sqrt{7} = 7 - 2\\sqrt{7} + 1 = (\\sqrt{7})^2 - 2\\cdot\\sqrt{7}\\cdot 1 + 1^2 = (\\sqrt{7}-1)^2$$ +Извлекаем корень. Так как $\\sqrt{7}\\approx 2{,}65\\gt 1$, выражение $\\sqrt{7}-1\\gt 0$: +$$\\sqrt{8-2\\sqrt{7}} = \\sqrt{(\\sqrt{7}-1)^2} = \\sqrt{7}-1$$ +Шаг 2. Преобразуем второе подкоренное выражение: $32 - 10\\sqrt{7}$. Здесь $2\\sqrt{mn}=10\\sqrt{7}$, то есть $\\sqrt{mn}=5\\sqrt{7}$, $mn=175$. И $m+n=32$. Подходят $m=25$, $n=7$: +$$32 - 10\\sqrt{7} = 25 - 10\\sqrt{7} + 7 = 5^2 - 2\\cdot 5\\cdot\\sqrt{7} + (\\sqrt{7})^2 = (5-\\sqrt{7})^2$$ +Так как $5\\gt\\sqrt{7}$, имеем $5-\\sqrt{7}\\gt 0$: +$$\\sqrt{32-10\\sqrt{7}} = 5-\\sqrt{7}$$ +Шаг 3. Складываем оба корня: +$$\\sqrt{8-2\\sqrt{7}} + \\sqrt{32-10\\sqrt{7}} = (\\sqrt{7}-1) + (5-\\sqrt{7}) = -1 + 5 + (\\sqrt{7}-\\sqrt{7}) = 4$$ +Шаг 4. Запишем число, обратное $4$: +$$\\dfrac{1}{4}$$ +
Ответ: $\\dfrac{1}{4}$
` + }, + { + text: `Периметр параллелограмма равен $30$ см, соседние стороны относятся как $2:3$. + Угол между высотами, проведёнными к соседним сторонам из вершины тупого угла + параллелограмма, равен $60^{\\circ}$. Найдите площадь параллелограмма.`, + sol: `Шаг 1 — стороны. +
Пусть $a=2k$, $b=3k$. Периметр: $2(2k+3k)=30 \\Rightarrow k=3$, т.е. $a=6$, $b=9$ см. +
Шаг 2 — острый угол. + + + + + + + + + + + + + + + 60° + A + B + C + D + $h_1$ + $h_2$ + +Из геометрии параллелограмма можно показать, что угол между двумя высотами из вершины тупого угла равен острому углу параллелограмма. Значит, острый угол $\\beta = 60°$. +
Шаг 3 — площадь. +$$S = a\\cdot b\\cdot\\sin\\beta = 6\\cdot 9\\cdot\\sin 60° = 54\\cdot\\frac{\\sqrt{3}}{2} = 27\\sqrt{3}\\text{ см}^2$$ +
Ответ: $27\\sqrt{3}$ см²
` + }, + { + text: `Во время строительных работ на Всебелорусской молодёжной стройке в Хатыни + двое студентов были определены на подсобные работы. Работая с одной скоростью, + они выполнили половину отведённой работы, затем увеличили скорость: один — на $20\\%$, + а второй — на $16\\%$, и вторую половину работы выполнили на один день раньше + запланированного времени. Успеют ли студенты выполнить работу за $14$ дней? + Ответ обоснуйте.`, + sol: `Пусть $n$ — плановое число дней. Скорость каждого $s = \\dfrac{1}{2n}$ ед/день, вместе: $\\dfrac{1}{n}$. +
Первая половина (работа $=\\tfrac{1}{2}$) — скорость не менялась: +$$t_1 = \\frac{1/2}{1/n} = \\frac{n}{2}\\text{ дней}$$ +Вторая половина — скорости увеличились: +$$\\text{1-й: }1{,}2s,\\quad \\text{2-й: }1{,}16s,\\quad \\text{вместе: }2{,}36s = \\frac{2{,}36}{2n} = \\frac{1{,}18}{n}$$ +$$t_2 = \\frac{1/2}{1{,}18/n} = \\frac{n}{2{,}36} = \\frac{25n}{59}\\text{ дней}$$ +Фактическое время: +$$t = t_1+t_2 = \\frac{n}{2}+\\frac{25n}{59} = \\frac{59n+50n}{118} = \\frac{109n}{118}$$ +Условие «на 1 день раньше»: +$$n - \\frac{109n}{118} = \\frac{9n}{118} = 1 \\implies n = \\frac{118}{9} \\approx 13{,}1\\text{ дней}$$ +Фактическое время выполнения: +$$t = \\frac{109n}{118} = \\frac{109}{9} = 12\\frac{1}{9} \\approx 12{,}1\\text{ дней}$$ +Так как $12{,}1 < 14$, студенты успеют выполнить работу за $14$ дней — с запасом около $2$ дней. +
Ответ: да, успеют ($\\approx 12{,}1$ дней $< 14$ дней)
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v14.js b/frontend/js/exam9/variants/v14.js new file mode 100644 index 0000000..e604619 --- /dev/null +++ b/frontend/js/exam9/variants/v14.js @@ -0,0 +1,234 @@ +VARIANTS[14] = { + label: "Вариант 14", + tasks: [ + { + text: `Определите, какая из следующих пар чисел является вершиной параболы + $y = -(x-2)^2 - 1$:`, + opts: [ + ["а", "$(2;\\ 1)$"], ["б", "$(2;\\ {-1})$"], ["в", "$(-2;\\ 1)$"], + ["г", "$(-2;\\ {-1})$"], ["д", "$(-1;\\ {-2})$"], + ], + sol: `Парабола $y = a(x-h)^2 + k$ имеет вершину $(h;\\,k)$. Здесь $h=2$, $k=-1$. + + + + + xy + + + (2; −1) + + 2 + −1 + +
Ответ: б) $(2;\\,-1)$
` + }, + { + text: `Определите, какие из чисел НЕ являются решением двойного неравенства + $-6 < 3x \\leq 12$:`, + opts: [ + ["а", "$-2$"], ["б", "$0$"], ["в", "$4$"], ["г", "$-3$"], ["д", "$1$"], + ], + sol: `Разделим на $3$: $-2 \\lt x \\leq 4$. +
Проверяем каждое число: +
    +
  • а) $-2$: не строго больше $-2$ ✗ — НЕ решение
    • +
  • б) $0$: $-2\\lt 0\\leq4$ ✓ — решение
    • +
  • в) $4$: $-2\\lt 4\\leq4$ ✓ — решение
    • +
  • г) $-3$: $-3\\lt-2$ ✗ — НЕ решение
    • +
  • д) $1$: ✓ — решение
    • +
+
Ответ: а) и г)
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "медианы треугольника пересекаются в одной точке;"], + ["б", "периметр равностороннего треугольника со стороной $a$ равен $3a$;"], + ["в", "основания равнобедренной трапеции равны;"], + ["г", "у прямоугольника все углы равны между собой?"], + ], + sol: `
    +
  • а) Медианы пересекаются в центроиде — верно
    • +
  • б) Периметр равностороннего треугольника $= 3a$ — верно
    • +
  • в) Основания равнобедренной трапеции равны — НЕВЕРНО
    • +
  • г) У прямоугольника все углы по $90°$ — верно
    • +
+В равнобедренной трапеции равны боковые стороны и углы при основаниях, но основания по определению различны по длине. +
Ответ: в)
` + }, + { + text: `Найдите значение выражения $4x - 9$, + если $\\dfrac{2x^2 - 7x}{x} = 0$.`, + sol: `При $x\\neq 0$: +$$\\frac{2x^2-7x}{x} = \\frac{x(2x-7)}{x} = 2x-7 = 0 \\implies x = \\frac{7}{2}$$ +$$4x-9 = 4\\cdot\\frac{7}{2}-9 = 14-9 = 5$$ +
Ответ: $5$
` + }, + { + text: `Диагональ $BD$ ромба $ABCD$ равна $24$ см. Угол между стороной $AB$ + и диагональю $AC$ равен $30^{\\circ}$. Найдите периметр ромба.`, + sol: ` + + + + + + + + + + + + 30° + + + 60° + + A + B + C + D + + a = ? + AC + BD = 24 + +Свойства ромба: все четыре стороны равны; диагонали — биссектрисы углов ромба. +
Признак равностороннего треугольника: равнобедренный треугольник с углом $60°$ является равносторонним. +

+Шаг 1. Диагональ $AC$ — биссектриса угла $A$ ромба, поэтому $\\angle BAC = \\dfrac{\\angle BAD}{2}$. По условию $\\angle BAC = 30°$, значит: +$$\\angle BAD = 2\\cdot 30° = 60°$$ +Шаг 2. Рассмотрим треугольник $ABD$: +
    +
  • $AB = AD = a$ (стороны ромба равны),
    • +
  • $\\angle BAD = 60°$.
    • +
+Равнобедренный треугольник с углом при вершине $60°$ — равносторонний. Значит, $BD = AB = a$. +
Шаг 3. По условию $BD = 24$ см, поэтому $a = 24$ см. +
Шаг 4. Периметр ромба: +$$P = 4a = 4\\cdot 24 = 96\\text{ см}$$ +
Ответ: $P = 96$ см
` + }, + { + text: `Сократите дробь $\\dfrac{25m^2 - 4n^2}{4n^2 + 25m^2 + 20mn}$.`, + sol: `Формула разности квадратов: $a^2-b^2 = (a-b)(a+b)$. +
Формула квадрата суммы: $(a+b)^2 = a^2+2ab+b^2$. +

+Шаг 1. Разложим числитель по формуле разности квадратов ($a=5m$, $b=2n$): +$$25m^2 - 4n^2 = (5m)^2 - (2n)^2 = (5m-2n)(5m+2n)$$ +Шаг 2. Разложим знаменатель. Это квадрат суммы: +$$25m^2 + 20mn + 4n^2 = (5m)^2 + 2\\cdot 5m\\cdot 2n + (2n)^2 = (5m+2n)^2$$ +Шаг 3. Сокращаем общий множитель $(5m+2n)$: +$$\\dfrac{(5m-2n)(5m+2n)}{(5m+2n)^2} = \\dfrac{5m-2n}{5m+2n}$$ +
Ответ: $\\dfrac{5m-2n}{5m+2n}$
` + }, + { + text: `Решите систему уравнений + $$\\begin{cases} x - 3y = 4, \\\\[4pt] xy - 7y = 6 \\end{cases}$$ + и найдите значение выражения $x_1 \\cdot y_1 + x_2 \\cdot y_2$, + где $(x_1;\\, y_1)$, $(x_2;\\, y_2)$ — решения системы.`, + sol: `Метод подстановки: выражаем одну переменную через другую и подставляем. +
Теорема Виета (обратная): $y^2+py+q=(y-y_1)(y-y_2)$, где $y_1+y_2=-p$, $y_1\\cdot y_2=q$. +

+Шаг 1. Из первого уравнения выразим $x$: +$$x = 4 + 3y$$ +Шаг 2. Подставим во второе: +$$(4+3y)y - 7y = 6$$ +$$4y + 3y^2 - 7y = 6$$ +$$3y^2 - 3y - 6 = 0$$ +Шаг 3. Разделим на $3$: +$$y^2 - y - 2 = 0$$ +Шаг 4. По теореме Виета: $y_1+y_2=1$, $y_1\\cdot y_2=-2$. Подходят $2$ и $-1$: +$$(y-2)(y+1)=0$$ +Шаг 5. Находим $x = 4+3y$ для каждого $y$: +
$y_1=2$:$x_1=4+6=10$
$y_2=-1$:$x_2=4-3=1$
+Шаг 6. Вычисляем требуемое выражение: +$$x_1y_1+x_2y_2 = 10\\cdot 2 + 1\\cdot(-1) = 20-1 = 19$$ +
Ответ: $19$
` + }, + { + text: `Найдите значение выражения $\\sqrt{19 - 6\\sqrt{10}} + \\sqrt{26 - 8\\sqrt{10}}$. + В ответ запишите число, противоположное полученному.`, + sol: `Формула квадрата разности: $(a-b)^2 = a^2 - 2ab + b^2$. +
Формула выноса корня из квадрата: $\\sqrt{a^2}=|a|$. Если $a\\geq 0$, то $|a|=a$; если $a\\lt 0$, то $|a|=-a$. +
Идея: подкоренное выражение вида $A - 2\\sqrt{B}$ часто можно представить как квадрат разности $(\\sqrt{m}-\\sqrt{n})^2$, где $m+n=A$ и $mn=B$. +

+Шаг 1. Преобразуем первое подкоренное выражение: $19 - 6\\sqrt{10}$. +
Здесь $2\\sqrt{mn}=6\\sqrt{10}$, то есть $\\sqrt{mn}=3\\sqrt{10}$, $mn=90$. И $m+n=19$. Подходят $m=9$, $n=10$: +$$19 - 6\\sqrt{10} = 9 - 6\\sqrt{10} + 10 = 3^2 - 2\\cdot 3\\cdot\\sqrt{10} + (\\sqrt{10})^2 = (3-\\sqrt{10})^2$$ +Извлекаем корень. Оценим $\\sqrt{10}$: так как $9\\lt 10\\lt 16$, имеем $3\\lt\\sqrt{10}\\lt 4$, точнее $\\sqrt{10}\\approx 3{,}16$. Значит, $3-\\sqrt{10}\\lt 0$, и модуль раскрывается как $|3-\\sqrt{10}|=\\sqrt{10}-3$: +$$\\sqrt{19-6\\sqrt{10}} = |3-\\sqrt{10}| = \\sqrt{10}-3$$ +Шаг 2. Преобразуем второе подкоренное выражение: $26 - 8\\sqrt{10}$. Здесь $2\\sqrt{mn}=8\\sqrt{10}$, $mn=160$, $m+n=26$. Подходят $m=16$, $n=10$: +$$26 - 8\\sqrt{10} = 16 - 8\\sqrt{10} + 10 = 4^2 - 2\\cdot 4\\cdot\\sqrt{10} + (\\sqrt{10})^2 = (4-\\sqrt{10})^2$$ +Так как $\\sqrt{10}\\approx 3{,}16\\lt 4$, имеем $4-\\sqrt{10}\\gt 0$: +$$\\sqrt{26-8\\sqrt{10}} = 4-\\sqrt{10}$$ +Шаг 3. Складываем оба корня: +$$(\\sqrt{10}-3) + (4-\\sqrt{10}) = -3 + 4 + (\\sqrt{10}-\\sqrt{10}) = 1$$ +Шаг 4. Запишем число, противоположное $1$: +$$-1$$ +
Ответ: $-1$
` + }, + { + text: `Периметр параллелограмма равен $32$ см, соседние стороны относятся как $3:5$. + Угол между высотами, проведёнными к соседним сторонам из вершины тупого угла + параллелограмма, равен $30^{\\circ}$. Найдите площадь параллелограмма.`, + sol: `Формула периметра параллелограмма: $P = 2(a+b)$. +
Формула площади параллелограмма: $S = a\\cdot b\\cdot\\sin\\alpha$, где $\\alpha$ — угол между сторонами. +
Свойство: угол между двумя высотами, проведёнными из вершины тупого угла параллелограмма к соседним сторонам, равен острому углу параллелограмма. +

+Шаг 1 — стороны. +
Пусть соседние стороны $a = 3k$ и $b = 5k$ (по условию относятся как $3:5$). Из формулы периметра: +$$2(3k + 5k) = 32$$ +$$16k = 32 \\implies k = 2$$ +Значит, $a = 6$ см, $b = 10$ см. +
Шаг 2 — острый угол параллелограмма. + + + + + + + + + + + + 30° + A + B + C + D + $h_1$ + $h_2$ + +Так как угол между двумя высотами из вершины тупого угла равен острому углу параллелограмма, получаем $\\beta = 30°$. +
Шаг 3 — площадь. +
По формуле площади параллелограмма: +$$S = a\\cdot b\\cdot\\sin\\beta = 6\\cdot 10\\cdot\\sin 30° = 60\\cdot\\dfrac{1}{2} = 30\\text{ см}^2$$ +
Ответ: $30$ см²
` + }, + { + text: `Во время строительных работ на Всебелорусской молодёжной стройке в Брестской + крепости двое студентов были определены на подсобные работы. Работая с одной + скоростью, они выполнили половину отведённой работы, затем увеличили скорость: + один — на $25\\%$, а второй — на $30\\%$, и вторую половину работы выполнили + на один день раньше запланированного времени. Успеют ли студенты выполнить + работу за $7$ дней? Ответ обоснуйте.`, + sol: `Пусть $n$ — плановое число дней. Скорость каждого $s = \\dfrac{1}{2n}$, вместе: $\\dfrac{1}{n}$. +
Первая половина: +$$t_1 = \\frac{1/2}{1/n} = \\frac{n}{2}\\text{ дней}$$ +Вторая половина — скорости увеличились: +$$\\text{1-й: }1{,}25s,\\quad \\text{2-й: }1{,}30s,\\quad \\text{вместе: }2{,}55s = \\frac{1{,}275}{n}$$ +$$t_2 = \\frac{1/2}{1{,}275/n} = \\frac{n}{2{,}55} = \\frac{20n}{51}\\text{ дней}$$ +Фактическое время: +$$t = \\frac{n}{2}+\\frac{20n}{51} = \\frac{51n+40n}{102} = \\frac{91n}{102}$$ +Условие «на 1 день раньше»: +$$n - \\frac{91n}{102} = \\frac{11n}{102} = 1 \\implies n = \\frac{102}{11} \\approx 9{,}3$$ +Фактическое время: +$$t = \\frac{91}{11} = 8\\frac{3}{11} \\approx 8{,}3\\text{ дней}$$ +Так как $8{,}3 \\gt 7$, студенты не успеют за $7$ дней. +
Ответ: нет, не успеют ($\\approx 8{,}3$ дней $\\gt 7$ дней)
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v15.js b/frontend/js/exam9/variants/v15.js new file mode 100644 index 0000000..d541240 --- /dev/null +++ b/frontend/js/exam9/variants/v15.js @@ -0,0 +1,223 @@ +VARIANTS[15] = { + label: "Вариант 15", + tasks: [ + { + text: `Определите, какое из следующих множеств может быть областью определения + чётной функции:`, + opts: [ + ["а", "$(-7;\\ 7]$"], + ["б", "$[-9;\\ {-1}) \\cup (-1;\\ 9]$"], + ["в", "$[-10;\\ 10]$"], + ["г", "$[-8;\\ {-1}) \\cup (-1;\\ 1) \\cup (1;\\ 8]$"], + ["д", "$[-11;\\ 11]$"], + ], + sol: `Область определения чётной функции должна быть симметрична относительно нуля: если $x$ входит, то и $-x$ входит. +
    +
  • а) $(-7;7]$: $x=7$ входит, $-7$ не входит (открытый конец) ✗
    • +
  • б) $[-9;{-1})\\cup(-1;9]$: $x=1$ входит, $-1$ не входит (исключена из обоих) ✗
    • +
  • в) $[-10;10]$: симметрично ✓
    • +
  • г) $[-8;{-1})\\cup(-1;1)\\cup(1;8]$: $\\pm1$ оба исключены симметрично, $\\pm8$ оба включены ✓
    • +
  • д) $[-11;11]$: симметрично ✓
    • +
+
Ответ: в), г) и д) — все три симметричны; наиболее простой пример: в)
` + }, + { + text: `Запись числового выражения $3^4 : 3^3 \\cdot 3^5$ в виде степени с основанием $3$ + имеет вид:`, + opts: [ + ["а", "$3^{12}$"], ["б", "$3^{-4}$"], ["в", "$3^6$"], + ["г", "$3^2$"], ["д", "$3^5$"], + ], + sol: `Деление — вычитаем показатели, умножение — складываем: +$$3^4 : 3^3 \\cdot 3^5 = 3^{4-3} \\cdot 3^5 = 3^1 \\cdot 3^5 = 3^{1+5} = 3^6$$ +
Ответ: в) $3^6$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "медианы треугольника делятся точкой пересечения в отношении $2:1$, считая от вершины;"], + ["б", "$\\cos 120^{\\circ} = \\dfrac{1}{2}$;"], + ["в", "диаметр окружности, вписанной в квадрат, равен его стороне;"], + ["г", "биссектриса любого угла делит этот угол на два равных угла?"], + ], + sol: `
    +
  • а) Медианы делятся в отношении $2:1$ — верно
    • +
  • б) $\\cos120°=\\frac{1}{2}$ — НЕВЕРНО: $\\cos120° = -\\frac{1}{2}$
    • +
  • в) Диаметр вписанной в квадрат окружности $= a$ (сторона) — верно
    • +
  • г) Биссектриса делит угол пополам — верно
    • +
+
Ответ: б)
` + }, + { + text: `Найдите третий член последовательности, заданной формулой $a_n = n^2 + 6n + 1$.`, + sol: `$$a_3 = 3^2 + 6\\cdot 3 + 1 = 9 + 18 + 1 = 28$$ +
Ответ: $28$
` + }, + { + text: `Четырёхугольник $ABCD$ вписан в окружность, $P$ — точка пересечения диагоналей, + $\\angle ADB = 72^{\\circ}$, $\\angle CBD = 64^{\\circ}$. Найдите $\\angle APB$.`, + sol: `Теорема о вписанных углах: вписанный угол равен половине дуги, на которую опирается. + + + + + + + P + A + B + C + D + 72° + 64° + +$\\angle ADB = 72°$ — вписанный угол, опирается на дугу $AB$: +$$\\overset{\\frown}{AB} = 2\\cdot72° = 144°$$ +$\\angle CBD = 64°$ — вписанный угол, опирается на дугу $CD$: +$$\\overset{\\frown}{CD} = 2\\cdot64° = 128°$$ +Угол между хордами равен полусумме перехваченных дуг: +$$\\angle APB = \\frac{\\overset{\\frown}{AB} + \\overset{\\frown}{CD}}{2} = \\frac{144° + 128°}{2} = \\frac{272°}{2} = 136°$$ +
Ответ: $136°$
` + }, + { + text: `Упростите выражение $|x - 3| + |x + 2| - 3$, если $x \\in (-2;\\ 0]$.`, + sol: `Определение модуля: $|A| = A$, если $A\\geq 0$, и $|A| = -A$, если $A\\lt 0$. +
Идея: чтобы раскрыть модуль, нужно определить знак подмодульного выражения на заданном промежутке. +

+Шаг 1. Определим знак $x-3$ при $x\\in(-2;\\,0]$. +
Так как $x\\leq 0$, то $x-3\\leq 0-3=-3\\lt 0$. Значит, $x-3\\lt 0$, и по определению модуля: +$$|x-3| = -(x-3) = 3-x$$ +Шаг 2. Определим знак $x+2$ при $x\\in(-2;\\,0]$. +
Так как $x\\gt -2$ (по условию), то $x+2\\gt 0$. По определению модуля: +$$|x+2| = x+2$$ +Шаг 3. Подставляем раскрытые модули в исходное выражение: +$$|x-3| + |x+2| - 3 = (3-x) + (x+2) - 3$$ +Шаг 4. Приводим подобные слагаемые: +$$3 - x + x + 2 - 3 = (3 + 2 - 3) + (-x + x) = 2$$ +Выражение оказалось постоянным на всём данном промежутке. +
Ответ: $2$ (константа на всём промежутке)
` + }, + { + text: `Найдите все значения переменной, при которых значение выражения + $\\dfrac{x + 3}{x + 2} - \\dfrac{2x}{x^2 - 4}$ равно нулю.`, + sol: `Решение дробно-рациональных уравнений: 1) находим ОДЗ; 2) умножаем обе части на общий знаменатель; 3) решаем полученное уравнение; 4) проверяем корни на принадлежность ОДЗ. +
Формула разности квадратов: $a^2-b^2 = (a-b)(a+b)$. +
Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. Запишем уравнение: +$$\\dfrac{x+3}{x+2} - \\dfrac{2x}{x^2-4} = 0$$ +Шаг 2. Разложим знаменатель $x^2-4$ по формуле разности квадратов: $x^2-4=(x-2)(x+2)$. Это и есть общий знаменатель. +
ОДЗ: $x\\neq 2$ и $x\\neq -2$. +
Шаг 3. Умножим обе части на $(x-2)(x+2)$: +$$(x+3)(x-2) - 2x = 0$$ +Шаг 4. Раскрываем скобки: +$$x^2 + 3x - 2x - 6 - 2x = 0$$ +$$x^2 - x - 6 = 0$$ +Шаг 5. По теореме Виета: $x_1+x_2=1$, $x_1\\cdot x_2=-6$. Подходят $3$ и $-2$: +$$(x-3)(x+2) = 0 \\implies x = 3 \\text{ или } x = -2$$ +Шаг 6. Проверяем ОДЗ: $x=-2$ не входит (отбрасываем). Остаётся $x=3$. +
Проверка $x=3$: +$$\\dfrac{3+3}{3+2} - \\dfrac{2\\cdot 3}{9-4} = \\dfrac{6}{5} - \\dfrac{6}{5} = 0 \\checkmark$$ +
Ответ: $x=3$
` + }, + { + text: `Если двузначное число разделить на сумму его цифр, в частном получим $7$, + а в остатке — $6$. Если число, составленное из тех же цифр в обратном порядке, + разделить на произведение цифр, то в частном получим $1$, а в остатке — $14$. + Найдите это число.`, + sol: `Пусть число $=10a+b$. +
Условие 1: $10a+b = 7(a+b)+6 \\Rightarrow 3a=6b+6 \\Rightarrow a=2b+2$ +
Условие 2: $10b+a = 1\\cdot ab+14$ (остаток $< $ делителя: $14 < ab$). +
Подставим $a=2b+2$: +$$10b+(2b+2) = (2b+2)b+14$$ +$$12b+2 = 2b^2+2b+14$$ +$$2b^2-10b+12=0 \\implies b^2-5b+6=0 \\implies (b-2)(b-3)=0$$ +$b=2$: $a=6$, число $62$. Проверим условие 2: $26\\div 12 = 2$ ост. $2\\neq 14$ ✗ (остаток должен быть $<$ делителя, $14>12$, значит $b=2$ не подходит). +
$b=3$: $a=8$, число $83$. +
Проверка: $83\\div 11=7$ ост. $6$ ✓; $38\\div 24=1$ ост. $14$ ✓ ($14<24$ ✓) +
Ответ: $83$
` + }, + { + text: `Найдите разность наибольшего и наименьшего целых решений двойного неравенства + $16 - 6x < x^2 \\leq 24 - 10x$.`, + sol: `Двойное неравенство $A \\lt B \\leq C$ равносильно системе $\\{A\\lt B,\\; B\\leq C\\}$, поэтому решаем каждую часть отдельно и находим пересечение. +
Метод интервалов для квадратного неравенства: раскладываем трёхчлен на множители и определяем знак на интервалах. +
Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. Решаем первую часть: $x^2 \\gt 16-6x$. +
Переносим всё влево: +$$x^2 + 6x - 16 \\gt 0$$ +По теореме Виета: $x_1+x_2=-6$, $x_1\\cdot x_2=-16$. Подходят $-8$ и $2$: +$$(x+8)(x-2) \\gt 0$$ +Произведение положительно вне корней: $x\\lt -8$ или $x\\gt 2$. +
Шаг 2. Решаем вторую часть: $x^2 \\leq 24-10x$. +
Переносим всё влево: +$$x^2 + 10x - 24 \\leq 0$$ +По теореме Виета: $x_1+x_2=-10$, $x_1\\cdot x_2=-24$. Подходят $-12$ и $2$: +$$(x+12)(x-2) \\leq 0$$ +Произведение неположительно между корнями: $-12\\leq x\\leq 2$. +
Шаг 3. Берём пересечение решений: +$$(x\\lt -8 \\text{ или } x\\gt 2)\\cap(-12\\leq x\\leq 2) = -12\\leq x \\lt -8$$ + + + + −12 + −10 + −8 + −6 + −4 + + + + +Шаг 4. Выписываем целые числа из $[-12;\\,-8)$: $-12,\\,-11,\\,-10,\\,-9$. +
Наибольшее $= -9$, наименьшее $= -12$. +
Шаг 5. Разность: +$$-9 - (-12) = 3$$ +
Ответ: $3$
` + }, + { + text: `Точка $M$ — середина стороны $CD$ параллелограмма $ABCD$ с площадью $360$ см². + Отрезок $BM$ пересекает диагональ $AC$ в точке $F$. + Найдите площадь четырёхугольника $AFMD$.`, + sol: `Шаг 1 — находим отношение AF:FC. +
Рассмотрим треугольники $ABF$ и $CMF$: +
    +
  • $AB\\parallel CM$ (т.к. $AB\\parallel DC$, а $CM$ — часть $DC$)
    • +
  • $\\angle FAB = \\angle FCM$ (накрест лежащие при параллельных)
    • +
  • $\\angle AFB = \\angle CFM$ (вертикальные углы)
    • +
+$\\Rightarrow$ треугольники подобны. Так как $AB = DC$ и $CM = DC/2$ (M — середина), то: +$$\\frac{AB}{CM} = \\frac{2}{1} \\Rightarrow \\frac{AF}{FC} = \\frac{AB}{CM} = 2 \\Rightarrow AF:FC = 2:1$$ +Шаг 2 — площадь треугольника ABM. +
Точка $M$ лежит на $DC\\parallel AB$, значит высота от $M$ до $AB$ равна высоте параллелограмма $h$. +$$S_{ABM} = \\tfrac{1}{2}\\cdot AB\\cdot h = \\tfrac{1}{2}\\cdot S_{ABCD} = 180$$ +Шаг 3 — площадь треугольника AMD. +
В треугольнике $ACD$ отрезок $AM$ — медиана (M — середина $CD$), поэтому: +$$S_{AMD} = \\tfrac{1}{2}\\cdot S_{ACD} = \\tfrac{1}{2}\\cdot 180 = 90$$ +Шаг 4 — площадь четырёхугольника ABMD. +$$S_{ABMD} = S_{ABM} + S_{AMD} = 180 + 90 = 270$$ +Шаг 5 — площадь треугольника ABF. +
Треугольники $ABF$ и $ABC$ имеют общую вершину $B$ и основания $AF$ и $AC$ на одной прямой: +$$\\frac{S_{ABF}}{S_{ABC}} = \\frac{AF}{AC} = \\frac{2}{3} \\Rightarrow S_{ABF} = \\frac{2}{3}\\cdot 180 = 120$$ +Шаг 6 — итог. +$$S_{AFMD} = S_{ABMD} - S_{ABF} = 270 - 120 = 150\\text{ см}^2$$ +
Ответ: $150$ см²
`, + figure: ` + + + + + + + A + B + C + D + M + F + AFMD +` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v16.js b/frontend/js/exam9/variants/v16.js new file mode 100644 index 0000000..f90032a --- /dev/null +++ b/frontend/js/exam9/variants/v16.js @@ -0,0 +1,231 @@ +VARIANTS[16] = { + label: "Вариант 16", + tasks: [ + { + text: `Определите, какое из следующих множеств НЕ может быть областью определения + нечётной функции:`, + opts: [ + ["а", "$(-\\infty;\\ {+\\infty})$"], + ["б", "$[-9;\\ 0) \\cup (0;\\ 9]$"], + ["в", "$[-10;\\ 10]$"], + ["г", "$(-8;\\ {-1}) \\cup (-1;\\ 1) \\cup (1;\\ 8)$"], + ["д", "$[-11;\\ 11)$"], + ], + sol: `Область нечётной функции симметрична относительно нуля. +
    +
  • а–г) все симметричны ✓
    • +
  • д) $[-11;11)$: $-11$ включён, $11$ исключён — несимметрично ✗
    • +
+
Ответ: д)
` + }, + { + text: `Запись числового выражения $3^4 \\cdot 3^3 : 3^2$ в виде степени с основанием $3$ + имеет вид:`, + opts: [ + ["а", "$3^9$"], ["б", "$3^{-1}$"], ["в", "$3^3$"], + ["г", "$3^4$"], ["д", "$3^5$"], + ], + sol: `Умножение — складываем показатели, деление — вычитаем: +$$3^4\\cdot3^3:3^2 = 3^{4+3-2} = 3^5$$ +
Ответ: д) $3^5$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "биссектрисы треугольника пересекаются в одной точке;"], + ["б", "$\\sin 150^{\\circ} = -\\dfrac{1}{2}$;"], + ["в", "диаметр окружности, описанной около квадрата, равен его диагонали;"], + ["г", "медиана треугольника делит сторону, к которой она проведена, пополам?"], + ], + sol: `
    +
  • а) Биссектрисы пересекаются в инцентре — верно
    • +
  • б) $\\sin150°=-\\dfrac{1}{2}$ — НЕВЕРНО: $\\sin150°=+\\dfrac{1}{2}$
    • +
  • в) Диаметр описанной окружности квадрата равен его диагонали — верно
    • +
  • г) Медиана проведена к середине стороны — верно
    • +
+
Ответ: б)
` + }, + { + text: `Найдите второй член последовательности, заданной формулой $a_n = 2n^2 + 5n + 1$.`, + sol: `$$a_2 = 2\\cdot2^2+5\\cdot2+1 = 8+10+1 = 19$$ +
Ответ: $19$
` + }, + { + text: `Четырёхугольник $ABCD$ вписан в окружность, $P$ — точка пересечения диагоналей, + $\\angle ACB = 48^{\\circ}$, $\\angle CAD = 54^{\\circ}$. Найдите $\\angle CPD$.`, + sol: ` + + + + + + P + A + B + C + D + 48° + 54° + +Теорема о вписанном угле: вписанный угол равен половине дуги, на которую он опирается; равносильно, дуга равна удвоенному вписанному углу. +
Теорема об угле между хордами: угол между двумя пересекающимися внутри окружности хордами равен полусумме двух перехваченных дуг. +

+Шаг 1. Угол $\\angle ACB = 48°$ — вписанный, опирается на дугу $AB$. По теореме о вписанном угле: +$$\\overset{\\frown}{AB} = 2\\cdot\\angle ACB = 2\\cdot 48° = 96°$$ +Шаг 2. Угол $\\angle CAD = 54°$ — вписанный, опирается на дугу $CD$: +$$\\overset{\\frown}{CD} = 2\\cdot\\angle CAD = 2\\cdot 54° = 108°$$ +Шаг 3. Точка $P$ — пересечение диагоналей $AC$ и $BD$ внутри окружности. Угол $\\angle CPD$ — между хордами $AC$ и $BD$, причём он перехватывает дуги $CD$ и $AB$ (противоположные дуги). +
Шаг 4. По теореме об угле между хордами: +$$\\angle CPD = \\dfrac{\\overset{\\frown}{CD}+\\overset{\\frown}{AB}}{2} = \\dfrac{108°+96°}{2} = \\dfrac{204°}{2} = 102°$$ +
Ответ: $102°$
` + }, + { + text: `Упростите выражение $|x - 4| + |x + 4| - 2$, если $x \\in (-4;\\ 0]$.`, + sol: `Определение модуля: $|A| = A$, если $A\\geq 0$, и $|A| = -A$, если $A\\lt 0$. +
Идея: для раскрытия модуля нужно определить знак подмодульного выражения на заданном промежутке. +

+Шаг 1. Определим знак $x-4$ при $x\\in(-4;\\,0]$. +
Так как $x\\leq 0$, то $x-4\\leq -4\\lt 0$. По определению модуля: +$$|x-4| = -(x-4) = 4-x$$ +Шаг 2. Определим знак $x+4$ на промежутке $(-4;\\,0]$. +
Так как $x\\gt -4$ (по условию), то $x+4\\gt 0$. По определению модуля: +$$|x+4| = x+4$$ +Шаг 3. Подставляем раскрытые модули в исходное выражение: +$$|x-4| + |x+4| - 2 = (4-x) + (x+4) - 2$$ +Шаг 4. Приводим подобные слагаемые: +$$4 - x + x + 4 - 2 = (4+4-2) + (-x+x) = 6$$ +Выражение оказалось постоянным на всём данном промежутке. +
Ответ: $6$ (константа на всём промежутке)
` + }, + { + text: `Найдите все значения переменной, при которых значение выражения + $\\dfrac{x + 2}{x + 3} + \\dfrac{2x}{x^2 - 9}$ равно нулю.`, + sol: `Решение дробно-рациональных уравнений: 1) находим ОДЗ; 2) умножаем обе части на общий знаменатель; 3) решаем полученное уравнение; 4) проверяем корни на принадлежность ОДЗ. +
Формула разности квадратов: $a^2-b^2 = (a-b)(a+b)$. +
Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. Запишем уравнение: +$$\\dfrac{x+2}{x+3} + \\dfrac{2x}{x^2-9} = 0$$ +Шаг 2. Разложим знаменатель $x^2-9$ по формуле разности квадратов: $x^2-9=(x-3)(x+3)$. Это общий знаменатель. +
ОДЗ: $x\\neq 3$ и $x\\neq -3$. +
Шаг 3. Умножим обе части уравнения на $(x-3)(x+3)$: +$$(x+2)(x-3) + 2x = 0$$ +Шаг 4. Раскрываем скобки: +$$x^2 + 2x - 3x - 6 + 2x = 0$$ +$$x^2 + x - 6 = 0$$ +Шаг 5. По теореме Виета: $x_1+x_2=-1$, $x_1\\cdot x_2=-6$. Подходят $-3$ и $2$: +$$(x+3)(x-2) = 0 \\implies x = -3 \\text{ или } x = 2$$ +Шаг 6. Проверяем ОДЗ: $x=-3$ не входит (отбрасываем). Остаётся $x=2$. +
Проверка $x=2$: +$$\\dfrac{2+2}{2+3} + \\dfrac{2\\cdot 2}{4-9} = \\dfrac{4}{5} + \\dfrac{4}{-5} = \\dfrac{4}{5} - \\dfrac{4}{5} = 0 \\checkmark$$ +
Ответ: $x=2$
` + }, + { + text: `Если двузначное число разделить на сумму его цифр, в частном получим $6$, + а в остатке — $5$. Если число, записанное теми же цифрами в обратном порядке, + разделить на произведение цифр, то в частном получим $2$, а в остатке — $5$. + Найдите это число.`, + sol: `Запись двузначного числа: $10a+b$, где $a$ — цифра десятков ($1\\leq a\\leq 9$), $b$ — цифра единиц ($0\\leq b\\leq 9$). Число с теми же цифрами в обратном порядке: $10b+a$. +
Теорема о делении с остатком: если $N$ при делении на $d$ даёт частное $q$ и остаток $r$, то $N = d\\cdot q + r$, $0\\leq r\\lt d$. +

+Шаг 1. Обозначим за $10a+b$ исходное число. +
Шаг 2. Первое условие: деление на сумму цифр $a+b$ даёт частное $6$ и остаток $5$: +$$10a + b = 6(a+b) + 5$$ +$$10a + b = 6a + 6b + 5$$ +$$4a - 5b = 5 \\quad (*)$$ +Шаг 3. Найдём пары цифр $(a;b)$, удовлетворяющие $(*)$. Преобразуем: +$$4a = 5b + 5 = 5(b+1)$$ +Левая часть делится на $5$, поэтому $a$ кратно $5$, то есть $a=5$ (другой кратный $5$ — это $0$, что невозможно для двузначного числа). +
При $a=5$: $20 = 5(b+1) \\Rightarrow b+1 = 4 \\Rightarrow b = 3$. +
Получаем число $\\boldsymbol{53}$. +
Шаг 4. Второе условие (проверка): число в обратном порядке $= 10b+a = 35$. Произведение цифр $= 5\\cdot 3 = 15$. По условию частное $2$, остаток $5$: +$$2\\cdot 15 + 5 = 30 + 5 = 35 \\checkmark$$ +И остаток $5\\lt 15$ — корректно. +
Проверка условия 1: сумма цифр $= 8$; $53:8 = 6$ (ост. $5$), так как $6\\cdot 8+5 = 48+5=53$ ✓. +
Ответ: $53$
` + }, + { + text: `Найдите разность наибольшего и наименьшего целых решений двойного неравенства + $x + 6 \\leq x^2 < 24 - 5x$.`, + sol: `Двойное неравенство $A\\leq B\\lt C$ равносильно системе $\\{A\\leq B,\\; B\\lt C\\}$. Решаем каждую часть и берём пересечение. +
Метод интервалов для квадратного неравенства: раскладываем трёхчлен на множители. +
Теорема Виета (обратная): $x^2+px+q=(x-x_1)(x-x_2)$, где $x_1+x_2=-p$, $x_1\\cdot x_2=q$. +

+Шаг 1. Решаем первую часть: $x+6 \\leq x^2$, то есть $x^2 \\geq x+6$. +
Переносим всё влево: +$$x^2 - x - 6 \\geq 0$$ +По теореме Виета: $x_1+x_2=1$, $x_1\\cdot x_2=-6$. Подходят $3$ и $-2$: +$$(x-3)(x+2) \\geq 0$$ +Произведение неотрицательно вне корней: $x\\leq -2$ или $x\\geq 3$. +
Шаг 2. Решаем вторую часть: $x^2 \\lt 24-5x$. +
Переносим всё влево: +$$x^2 + 5x - 24 \\lt 0$$ +По теореме Виета: $x_1+x_2=-5$, $x_1\\cdot x_2=-24$. Подходят $-8$ и $3$: +$$(x+8)(x-3) \\lt 0$$ +Произведение отрицательно между корнями: $-8\\lt x\\lt 3$. +
Шаг 3. Берём пересечение: +$$(x\\leq -2 \\text{ или } x\\geq 3) \\cap (-8\\lt x\\lt 3) = -8 \\lt x \\leq -2$$ + + + + −9 + −7 + −5 + −3 + −1 + + + + +Шаг 4. Целые числа из $(-8;\\,-2]$: $-7,\\,-6,\\,-5,\\,-4,\\,-3,\\,-2$. +
Наибольшее $= -2$, наименьшее $= -7$. +
Шаг 5. Разность: +$$-2 - (-7) = 5$$ +
Ответ: $5$
` + }, + { + text: `Точка $M$ — середина стороны $BC$ параллелограмма $ABCD$ с площадью $240$ см². + Отрезок $AM$ пересекает диагональ $BD$ в точке $F$. + Найдите площадь четырёхугольника $FMCD$.`, + figure: ` + + + + + + + A + B + C + D + M + F + FMCD +`, + sol: `Метод координат: вводим координаты вершин параллелограмма, чтобы свести задачу к вычислениям. Площадь от выбора координат не зависит — отношение площадей сохраняется. +
Формула Гаусса (площадь многоугольника по координатам): для четырёхугольника с вершинами $(x_i;y_i)$: +$$2S = \\left|\\sum_i x_i(y_{i+1}-y_{i-1})\\right|$$ +

+Шаг 1 — выбираем координаты. +
Поместим вершины параллелограмма так: $A=(0;0)$, $B=(1;0)$, $C=(1;1)$, $D=(0;1)$ (получится квадрат единичной площади, но отношения площадей такие же, как в любом параллелограмме). +
$M$ — середина $BC$, поэтому $M=\\bigl(1;\\tfrac{1}{2}\\bigr)$. +
Шаг 2 — находим точку F. +
Прямая $AM$: точки вида $(t;\\tfrac{t}{2})$, где $t\\in[0;1]$. +
Прямая $BD$: точки вида $(1-s;s)$, где $s\\in[0;1]$. +
В точке $F$ обе прямые пересекаются: +$$t = 1-s,\\quad \\tfrac{t}{2} = s$$ +Подставляем $s = \\tfrac{t}{2}$ в первое: $t = 1-\\tfrac{t}{2} \\Rightarrow \\tfrac{3t}{2}=1 \\Rightarrow t=\\tfrac{2}{3}$, $s=\\tfrac{1}{3}$. +
Получаем $F=\\bigl(\\tfrac{2}{3};\\tfrac{1}{3}\\bigr)$. +
Шаг 3 — площадь FMCD по формуле Гаусса. +
Вершины (в порядке обхода): $F\\bigl(\\tfrac{2}{3};\\tfrac{1}{3}\\bigr)$, $M\\bigl(1;\\tfrac{1}{2}\\bigr)$, $C(1;1)$, $D(0;1)$. +
Применяем формулу: +$$2S = \\left|\\tfrac{2}{3}\\bigl(\\tfrac{1}{2}-1\\bigr)+1\\bigl(1-\\tfrac{1}{3}\\bigr)+1\\bigl(1-\\tfrac{1}{2}\\bigr)+0\\bigl(\\tfrac{1}{2}-1\\bigr)\\right|$$ +$$= \\left|-\\tfrac{1}{3}+\\tfrac{2}{3}+\\tfrac{1}{2}\\right| = \\tfrac{5}{6}$$ +Значит $S = \\tfrac{5}{12}$ от площади выбранного «единичного» параллелограмма. +
Шаг 4 — итог. +
Реальная площадь параллелограмма $= 240$ см², поэтому: +$$S_{FMCD} = \\tfrac{5}{12}\\cdot 240 = 100\\text{ см}^2$$ +
Ответ: $100$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v17.js b/frontend/js/exam9/variants/v17.js new file mode 100644 index 0000000..4e2b886 --- /dev/null +++ b/frontend/js/exam9/variants/v17.js @@ -0,0 +1,202 @@ +VARIANTS[17] = { + label: "Вариант 17", + tasks: [ + { + text: `Определите рисунок, на котором изображён график функции $y = \\sqrt{x} - 1$:`, + figure: ``, + sol: `Функция $y=\\sqrt{x}-1$: +
    +
  • Область определения: $x\\geq 0$ (начинается с нуля)
    • +
  • При $x=0$: $y=-1$ — начало кривой в точке $(0;\\,-1)$
    • +
  • При $x=1$: $y=0$ — пересекает ось $Ox$ в точке $(1;\\,0)$
    • +
  • Возрастает при $x>0$
    • +
+На рисунке ищем кривую, которая начинается НИЖЕ оси $Ox$ при $x=0$. +
Ответ: д)
` + }, + { + text: `Из данных чисел выберите те, которые НЕ входят в область определения выражения $\\dfrac{2}{\\sqrt{2x-4}}$:`, + opts: [ + ["а", "$\\dfrac{1}{2}$"], ["б", "$2{,}5$"], ["в", "$2$"], + ["г", "$3$"], ["д", "$4$"], + ], + sol: `Знаменатель не равен нулю и подкоренное выражение положительно: $2x-4>0 \\Rightarrow x>2$. +
ОДЗ: $x>2$. Проверяем: +
    +
  • а) $\\frac{1}{2}<2$ ✗ — НЕ входит
    • +
  • б) $2{,}5>2$ ✓ — входит
    • +
  • в) $2=2$ → знаменатель $=0$ ✗ — НЕ входит
    • +
  • г) $3>2$ ✓ — входит
    • +
  • д) $4>2$ ✓ — входит
    • +
+
Ответ: а) и в)
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "у правильного $n$-угольника все стороны равны между собой;"], + ["б", "по теореме косинусов для треугольника со сторонами $a$, $b$, $c$ верно, что $a^2 = b^2 + c^2 - 2bc\\cos\\alpha$;"], + ["в", "площадь ромба равна половине произведения диагоналей;"], + ["г", "площадь круга с радиусом $r$ равна $2\\pi r$?"], + ], + sol: `
    +
  • а) Правильный $n$-угольник — все стороны равны верно
    • +
  • б) Теорема косинусов $a^2=b^2+c^2-2bc\\cos\\alpha$ — верно
    • +
  • в) $S_{\\text{ромб}}=\\frac{d_1 d_2}{2}$ — верно
    • +
  • г) Площадь круга $=2\\pi r$ — НЕВЕРНО
    • +
+$S_{\\text{круга}} = \\pi r^2$, а $2\\pi r$ — это длина окружности. +
Ответ: г)
` + }, + { + text: `Найдите значение выражения $x \\cdot y$, где $(x;\\, y)$ — решение системы уравнений + $$\\begin{cases} x + y = 5, \\\\[4pt] 2x - y = 4. \\end{cases}$$`, + sol: `Сложим оба уравнения: $3x=9 \\Rightarrow x=3$. +
Из первого: $y=5-3=2$. +$$x\\cdot y = 3\\cdot 2 = 6$$ +
Ответ: $6$
` + }, + { + text: `Четырёхугольник $ABCD$ вписан в окружность, центр $O$ окружности лежит на стороне $AD$. + Найдите угол $CAD$, если угол $ABC$ равен $118^{\\circ}$.`, + sol: `Ключевой факт: $O$ лежит на $AD$ $\\Rightarrow$ $AD$ — диаметр окружности. + + + + + + + + + + + + O + + + + + 90° + + + + 28° + + + + 62° + + A + D + B + C + + 118° + +Шаг 1. $AD$ — диаметр $\\Rightarrow$ $\\angle ACD = 90°$ (вписанный угол, опирающийся на диаметр). +
Шаг 2. $ABCD$ — вписанный четырёхугольник, противоположные углы в сумме дают $180°$: +$$\\angle ADC = 180° - \\angle ABC = 180° - 118° = 62°$$ +Шаг 3. В треугольнике $ACD$: $\\angle ACD=90°$, $\\angle ADC=62°$: +$$\\angle CAD = 180° - 90° - 62° = 28°$$ +
Ответ: $\\angle CAD = 28°$
` + }, + { + text: `Найдите количество целых решений неравенства $x^2 + 5x < 14$.`, + sol: `Метод интервалов для квадратного неравенства: приводим к виду $ax^2+bx+c\\lt 0$, раскладываем на множители и определяем знаки. +
Шаг 1. Переносим всё в одну часть: $x^2+5x-14\\lt 0$. +
Шаг 2. Раскладываем левую часть на множители. Ищем два числа, произведение которых равно $-14$, а сумма $5$. Это $7$ и $-2$, значит $x^2+5x-14=(x+7)(x-2)$. +
Шаг 3. Решаем неравенство $(x+7)(x-2)\\lt 0$. Произведение двух множителей отрицательно, когда они разных знаков, поэтому $-7\\lt x\\lt 2$. + + + + −7 + −5 + −3 + −1 + 1 + 2 + + + + +Шаг 4. Выписываем целые числа из интервала $(-7;\\;2)$ — концы не входят, поскольку неравенство строгое: $-6,\\,-5,\\,-4,\\,-3,\\,-2,\\,-1,\\,0,\\,1$ — всего 8 чисел. +
Ответ: $8$
` + }, + { + text: `Найдите $55\\%$ от значения выражения $\\dfrac{71^2 - 23^2 + 94 \\cdot 42}{62^2 - 32^2}$.`, + sol: `Формула разности квадратов: $a^2-b^2=(a-b)(a+b)$. Правило процентов: $p\\%$ от числа $N$ равно $\\dfrac{p}{100}\\cdot N$. +
Шаг 1. Преобразуем числитель. По формуле разности квадратов: +$$71^2-23^2 = (71-23)(71+23) = 48\\cdot 94.$$ +Значит, в числителе $48\\cdot 94 + 94\\cdot 42$. Так как множитель $94$ общий, выносим его за скобки: +$$48\\cdot 94 + 94\\cdot 42 = 94\\cdot(48+42) = 94\\cdot 90 = 8460.$$ +Шаг 2. Преобразуем знаменатель по той же формуле: +$$62^2-32^2 = (62-32)(62+32) = 30\\cdot 94 = 2820.$$ +Шаг 3. Делим числитель на знаменатель. Замечаем, что общий множитель $94$ сокращается: +$$\\dfrac{94\\cdot 90}{30\\cdot 94} = \\dfrac{90}{30} = 3.$$ +Шаг 4. Находим $55\\%$ от полученного числа. По правилу процентов: +$$55\\%\\text{ от }3 = \\dfrac{55}{100}\\cdot 3 = 0{,}55\\cdot 3 = 1{,}65.$$ +
Ответ: $1{,}65$
` + }, + { + text: `Найдите значение выражения $a + x_0$, где $a$ — положительное число, + при котором левая часть уравнения $4x^2 - ax + 25 = 0$ является квадратом разности, + а $x_0$ — корень уравнения.`, + sol: `Формула квадрата разности: $(A-B)^2 = A^2 - 2AB + B^2$. +
Шаг 1. Чтобы $4x^2 - ax + 25$ был квадратом разности, надо представить его в виде $(A-B)^2$. Так как $4x^2=(2x)^2$ и $25=5^2$, берём $A=2x$ и $B=5$ (положительное, так как по условию $a\\gt 0$, а средний член имеет знак минус). +
Шаг 2. Сравниваем средние члены: $-ax$ и $-2AB = -2\\cdot 2x\\cdot 5 = -20x$. Значит $a=20$. +
Шаг 3. Подставляем $a=20$ и решаем уравнение: +$$4x^2-20x+25 = (2x-5)^2 = 0 \\;\\implies\\; 2x-5=0 \\;\\implies\\; x_0 = \\dfrac{5}{2}$$ +Шаг 4. Находим искомое выражение: +$$a+x_0 = 20+\\dfrac{5}{2} = 22{,}5$$ +
Ответ: $22{,}5$
` + }, + { + text: `В равнобедренной трапеции $ABCD$ с основаниями $AD$ и $BC$ проведена высота $BK$. + Найдите площадь трапеции, если $BK = 6$ см, диагональ $AC = 10$ см.`, + sol: `В равнобедренной трапеции $AD\\parallel BC$, высота $BK\\perp AD$. + + + + + + A + D + C + B + K + BK=6 + AC=10 + +Введём координаты: $A=(0,0)$, основание $AD$ по оси $Ox$, высота $h=BK=6$. +
Тогда $B=(p,\\ 6)$ и $C=(p+b,\\ 6)$, где $p=\\dfrac{AD-BC}{2}$ — горизонтальное смещение. +

+Диагональ $AC$ идёт от $A(0,0)$ до $C(p+b,\; 6)$: +$$AC^2 = (p+b)^2 + 6^2$$ +Заметим, что $p+b = \\dfrac{AD-BC}{2}+BC = \\dfrac{AD+BC}{2} = \\text{средняя линия}$. +
Из условия $AC=10$: +$$\\text{средняя линия} = \\sqrt{AC^2 - h^2} = \\sqrt{100-36} = \\sqrt{64} = 8$$ +$$S = \\text{средняя линия}\\times h = 8\\times 6 = 48\\text{ см}^2$$ +
Ответ: $48$ см²
` + }, + { + text: `В аквапарке «Лебяжий», который расположен в городе Минске, один из бассейнов + до максимальной метки можно заполнять через две трубы, причём через первую — + на $5$ часов дольше, чем через вторую. Заполнение бассейна через обе трубы + одновременно продолжается не менее $6$ часов. + За какое наименьшее количество часов можно заполнить бассейн через первую трубу?`, + sol: `Пусть первая труба заполняет за $t_1$ часов, вторая — за $t_2=t_1-5$ часов. +
Совместная скорость заполнения: $\\dfrac{1}{t_1}+\\dfrac{1}{t_1-5}$ (бассейнов в час). +
Условие «не менее 6 часов» вместе означает скорость $\\leq \\dfrac{1}{6}$: +$$\\frac{1}{t_1}+\\frac{1}{t_1-5} \\leq \\frac{1}{6}$$ +$$\\frac{2t_1-5}{t_1(t_1-5)} \\leq \\frac{1}{6}$$ +$$6(2t_1-5) \\leq t_1(t_1-5)$$ +$$12t_1-30 \\leq t_1^2-5t_1$$ +$$t_1^2-17t_1+30 \\geq 0$$ +$$(t_1-15)(t_1-2) \\geq 0$$ +$$t_1 \\leq 2 \\quad\\text{или}\\quad t_1 \\geq 15$$ +Так как $t_2=t_1-5>0$, необходимо $t_1>5$. Поэтому $t_1\\geq 15$. +
Проверка $t_1=15$: $t_2=10$, вместе: $\\frac{1}{15}+\\frac{1}{10}=\\frac{1}{6}$ (ровно 6 ч) ✓ +
Ответ: не менее $15$ часов
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v18.js b/frontend/js/exam9/variants/v18.js new file mode 100644 index 0000000..37bae8e --- /dev/null +++ b/frontend/js/exam9/variants/v18.js @@ -0,0 +1,197 @@ +VARIANTS[18] = { + label: "Вариант 18", + tasks: [ + { + text: `Определите рисунок, на котором изображён график функции $y = |x| + 1$:`, + figure: ``, + sol: `Функция $y=|x|+1$: +
    +
  • При $x=0$: $y=1$ — вершина V-образной фигуры в точке $(0;\\,1)$
    • +
  • Пересечение с осью $Ox$: $|x|+1=0 \\implies |x|=-1$ — нет пересечений (весь график выше оси $Ox$)
    • +
  • При $x>0$: $y=x+1$ (луч вправо-вверх); при $x<0$: $y=-x+1$ (луч влево-вверх)
    • +
  • График симметричен относительно оси $Oy$
    • +
+На рисунке ищем V-образную кривую с вершиной в точке $(0;\\,1)$, целиком выше оси $Ox$. +
Ответ: рисунок с V-образным графиком, вершина которого в точке $(0;\\,1)$
` + }, + { + text: `Из данных чисел выберите те, которые НЕ входят в область определения выражения $\\dfrac{2}{\\sqrt{3x-9}}$:`, + opts: [ + ["а", "$\\dfrac{1}{3}$"], ["б", "$3{,}5$"], ["в", "$3$"], + ["г", "$4$"], ["д", "$5$"], + ], + sol: `Знаменатель не равен нулю и подкоренное выражение положительно: $3x-9>0 \\Rightarrow x>3$. +
ОДЗ: $x>3$. Проверяем: +
    +
  • а) $\\frac{1}{3}<3$ ✗ — НЕ входит
    • +
  • б) $3{,}5>3$ ✓ — входит
    • +
  • в) $3=3$ → знаменатель $=0$ ✗ — НЕ входит
    • +
  • г) $4>3$ ✓ — входит
    • +
  • д) $5>3$ ✓ — входит
    • +
+
Ответ: а) и в)
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "у правильного $n$-угольника все углы равны;"], + ["б", "по теореме косинусов для треугольника со сторонами $a$, $b$, $c$ верно, что $b^2 = a^2 + c^2 - 2ac\\cos\\beta$;"], + ["в", "площадь прямоугольного треугольника равна половине произведения катетов;"], + ["г", "длина окружности с радиусом $R$ равна $\\pi R$?"], + ], + sol: `
    +
  • а) Правильный $n$-угольник — все углы равны — верно
    • +
  • б) Теорема косинусов $b^2=a^2+c^2-2ac\\cos\\beta$ — верно
    • +
  • в) $S=\\frac{1}{2}\\cdot\\text{катет}_1\\cdot\\text{катет}_2$ — верно
    • +
  • г) Длина окружности $=\\pi R$ — НЕВЕРНО
    • +
+Длина окружности равна $2\\pi R$, а не $\\pi R$. +
Ответ: г)
` + }, + { + text: `Найдите значение выражения $x \\cdot y$, где $(x;\\, y)$ — решение системы уравнений + $$\\begin{cases} x + y = 5, \\\\[4pt] 3x - y = 7. \\end{cases}$$`, + sol: `Сложим оба уравнения: $4x=12 \\Rightarrow x=3$. +
Из первого: $y=5-3=2$. +$$x\\cdot y = 3\\cdot 2 = 6$$ +
Ответ: $6$
` + }, + { + text: `Четырёхугольник $ABCD$ вписан в окружность, центр $O$ окружности лежит на стороне $AD$. + Найдите угол $BCD$, если угол $ADB$ равен $32^{\\circ}$.`, + sol: `Ключевой факт: $O$ лежит на $AD$ $\\Rightarrow$ $AD$ — диаметр окружности. + + + + + + + + + + + + O + + + 90° + + + 32° + + A + D + B + C + + 122° + +Шаг 1. $AD$ — диаметр $\\Rightarrow$ $\\angle ABD = 90°$ (вписанный угол, опирающийся на диаметр). +
Шаг 2. В треугольнике $ABD$: $\\angle ABD=90°$, $\\angle ADB=32°$: +$$\\angle BAD = 180° - 90° - 32° = 58°$$ +Шаг 3. $ABCD$ — вписанный четырёхугольник, противоположные углы в сумме дают $180°$: +$$\\angle BCD = 180° - \\angle BAD = 180° - 58° = 122°$$ +
Ответ: $\\angle BCD = 122°$
` + }, + { + text: `Найдите количество целых решений неравенства $x^2 + 4x < 12$.`, + sol: `Метод интервалов для квадратного неравенства: переносим всё в одну часть, раскладываем на множители и находим знаки. +
Шаг 1. Переносим $12$ влево: $x^2+4x-12\\lt 0$. +
Шаг 2. Раскладываем на множители. Ищем два числа с произведением $-12$ и суммой $4$ — это $6$ и $-2$, поэтому $x^2+4x-12=(x+6)(x-2)$. +
Шаг 3. Решаем $(x+6)(x-2)\\lt 0$. Произведение отрицательно, когда множители разных знаков, значит $-6\\lt x\\lt 2$. + + + + −6 + −4 + −2 + 0 + 2 + + + + +Шаг 4. Выписываем целые числа из интервала $(-6;\\;2)$, не включая концы (неравенство строгое): $-5,\\,-4,\\,-3,\\,-2,\\,-1,\\,0,\\,1$ — всего 7 чисел. +
Ответ: $7$
` + }, + { + text: `Найдите $75\\%$ от значения выражения $\\dfrac{62^2 - 12^2 + 74 \\cdot 46}{53^2 - 21^2}$.`, + sol: `Формула разности квадратов: $a^2-b^2=(a-b)(a+b)$. Правило процентов: $p\\%$ от числа $N$ равно $\\dfrac{p}{100}\\cdot N$. +
Шаг 1. Преобразуем числитель. По формуле разности квадратов: +$$62^2-12^2 = (62-12)(62+12) = 50\\cdot 74.$$ +Значит, в числителе $50\\cdot 74 + 74\\cdot 46$. Множитель $74$ общий, выносим его за скобки: +$$50\\cdot 74 + 74\\cdot 46 = 74\\cdot(50+46) = 74\\cdot 96 = 7104.$$ +Шаг 2. Преобразуем знаменатель аналогично: +$$53^2-21^2 = (53-21)(53+21) = 32\\cdot 74 = 2368.$$ +Шаг 3. Делим числитель на знаменатель — общий множитель $74$ сокращается: +$$\\dfrac{74\\cdot 96}{32\\cdot 74} = \\dfrac{96}{32} = 3.$$ +Шаг 4. Находим $75\\%$ от полученного числа: +$$75\\%\\text{ от }3 = \\dfrac{75}{100}\\cdot 3 = 0{,}75\\cdot 3 = 2{,}25.$$ +
Ответ: $2{,}25$
` + }, + { + text: `Найдите значение выражения $a + x_0$, где $a$ — отрицательное число, + при котором левая часть уравнения $4x^2 + ax + 9 = 0$ является квадратом разности, + а $x_0$ — корень уравнения.`, + sol: `Формула квадрата разности: $(A-B)^2 = A^2 - 2AB + B^2$. +
Шаг 1. Заметим, что $4x^2=(2x)^2$ и $9=3^2$. Значит, представляем выражение в виде $(2x-3)^2$ или $(2x+3)^2$. +
Шаг 2. Раскрываем квадраты: +
$(2x-3)^2 = 4x^2-12x+9$ — здесь средний коэффициент $-12$; +
$(2x+3)^2 = 4x^2+12x+9$ — здесь средний коэффициент $+12$. +
Шаг 3. По условию $a$ отрицательное, значит $a=-12$. +
Шаг 4. Решаем уравнение $4x^2-12x+9=(2x-3)^2=0$. Отсюда $2x-3=0$, и $x_0=\\dfrac{3}{2}$. +
Шаг 5. Находим $a+x_0$: +$$a+x_0 = -12+\\dfrac{3}{2} = -10{,}5$$ +
Ответ: $-10{,}5$
` + }, + { + text: `В равнобедренной трапеции $ABCD$ с основаниями $AD$ и $BC$ проведена высота $CH$. + Найдите площадь трапеции, если $CH = 12$ см, диагональ $BD = 15$ см.`, + sol: `Свойство равнобедренной трапеции: если из вершины меньшего основания опустить высоту на большее основание, то её основание отстоит от ближайшей вершины большего основания на $\\dfrac{AD-BC}{2}$. +
Формула площади трапеции: $S = \\dfrac{a+b}{2}\\cdot h$ — произведение средней линии на высоту. +
Теорема Пифагора: в прямоугольном треугольнике $c^2=a^2+b^2$. + + + + + + A + D + C + B + H + CH=12 + BD=15 + +Шаг 1. Введём координаты: $A=(0;0)$, $D=(AD;0)$, $C=(AD - p;\\,12)$, $B=(p;\\,12)$, где $p=\\dfrac{AD-BC}{2}$. Это даёт высоту $CH=12$, опущенную из $C$ в точку $H=(AD-p;\\,0)$. +
Шаг 2. Найдём $DH$ — отрезок от $D$ до основания высоты: +$$DH = AD - (AD-p) = p = \\dfrac{AD-BC}{2}$$ +Шаг 3. Замечаем: средняя линия трапеции равна $\\dfrac{AD+BC}{2}$, а длина $BH$ (горизонтальная проекция диагонали $BD$) как раз и равна $AD-p = \\dfrac{AD+BC}{2}$ — это средняя линия. +
Шаг 4. Треугольник $BHD$ прямоугольный (так как $CH\\perp AD$ и $BH\\parallel AD$). По теореме Пифагора: +$$\\text{средняя линия} = BH = \\sqrt{BD^2 - CH^2} = \\sqrt{15^2-12^2} = \\sqrt{225-144} = \\sqrt{81} = 9\\text{ см}$$ +Шаг 5. Применяем формулу площади: +$$S = \\text{средняя линия}\\cdot h = 9\\cdot 12 = 108\\text{ см}^2$$ +
Ответ: $108$ см²
` + }, + { + text: `В аквацентре, который расположен в городе Гродно, один из бассейнов можно заполнять + через две трубы, причём заполнение до максимальной метки через вторую — + на $5$ часов быстрее, чем через первую. Заполнение бассейна через обе трубы + одновременно продолжается не более $6$ часов. + За какое наибольшее количество часов можно заполнить бассейн через вторую трубу?`, + sol: `Пусть вторая труба заполняет за $t_2$ часов, тогда первая — за $t_1=t_2+5$ часов. +
Условие «не более 6 часов» означает, что совместная скорость $\\geq \\dfrac{1}{6}$: +$$\\frac{1}{t_1}+\\frac{1}{t_2} \\geq \\frac{1}{6}$$ +$$\\frac{1}{t_2+5}+\\frac{1}{t_2} \\geq \\frac{1}{6}$$ +$$\\frac{2t_2+5}{t_2(t_2+5)} \\geq \\frac{1}{6}$$ +$$6(2t_2+5) \\geq t_2(t_2+5)$$ +$$12t_2+30 \\geq t_2^2+5t_2$$ +$$t_2^2-7t_2-30 \\leq 0$$ +$$(t_2-10)(t_2+3) \\leq 0$$ +$$-3 \\leq t_2 \\leq 10$$ +Так как $t_2>0$, получаем $0 < t_2 \\leq 10$. +
Проверка $t_2=10$: $t_1=15$, вместе: $\\frac{1}{15}+\\frac{1}{10}=\\frac{1}{6}$ (ровно 6 ч) ✓ +
Ответ: не более $10$ часов
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v19.js b/frontend/js/exam9/variants/v19.js new file mode 100644 index 0000000..6f919ac --- /dev/null +++ b/frontend/js/exam9/variants/v19.js @@ -0,0 +1,200 @@ +VARIANTS[19] = { + label: "Вариант 19", + tasks: [ + { + text: `Определите рисунок, на котором изображён график функции $y = (x+1)^3$:`, + figure: ``, + sol: `Функция $y=(x+1)^3$ — это график $y=x^3$, сдвинутый на $1$ единицу влево. +
Ключевые свойства: +
    +
  • Точка перегиба: $(-1;\\,0)$ — на оси $Ox$ при $x=-1$
    • +
  • $y$-пересечение: при $x=0$, $y=1$ — точка $(0;\\,1)$
    • +
  • Возрастает на всей $\\mathbb{R}$ (классическая S-образная форма куба)
    • +
  • Проходит через $(-2;\\,-1)$, так как $(-2+1)^3=-1$
    • +
+
Ответ: график, пересекающий ось $Ox$ в точке $(-1;\\,0)$
` + }, + { + text: `Какие из данных чисел являются решениями системы неравенств + $$\\left\\{\\begin{array}{l} x < 7, \\\\[4pt] x \\geq -\\dfrac{1}{2}; \\end{array}\\right.$$`, + opts: [ + ["а", "$6$"], ["б", "$7$"], ["в", "$0$"], + ["г", "$-\\dfrac{3}{4}$"], ["д", "$-1$"], + ], + sol: `Решение системы: $-\\dfrac{1}{2}\\leq x<7$. + + + + −1 + 0 + 7 + + + + −1/2 + +Проверяем каждое: +
    +
  • а) $6$: $-\\frac{1}{2}\\leq 6<7$ ✓
    • +
  • б) $7$: условие $x<7$ нарушено ✗
    • +
  • в) $0$: $-\\frac{1}{2}\\leq 0<7$ ✓
    • +
  • г) $-\\frac{3}{4}=-0{,}75$: $-0{,}75 < -0{,}5$ ✗
    • +
  • д) $-1<-\\frac{1}{2}$ ✗
    • +
+
Ответ: а) и в)
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "диагонали параллелограмма точкой пересечения делятся пополам;"], + ["б", "площадь квадрата со стороной $4$ см равна $16$ см²;"], + ["в", "если четырёхугольник $ABCD$ вписан в окружность, то $\\angle A + \\angle B = 180^{\\circ}$;"], + ["г", "если у $\\triangle ABC$ $\\angle C = 90^{\\circ}$, $AC = 3$, $BC = 4$, то $\\operatorname{tg} A = \\dfrac{4}{3}$?"], + ], + sol: `
    +
  • а) Диагонали параллелограмма делятся пополам — верно
    • +
  • б) $S_{\\text{кв}}=4^2=16$ см² — верно
    • +
  • в) Во вписанном четырёхугольнике $\\angle A+\\angle B=180°$ — НЕВЕРНО
    • +
  • г) $\\operatorname{tg}A = \\dfrac{BC}{AC}=\\dfrac{4}{3}$ — верно
    • +
+В вписанном четырёхугольнике в сумме $180°$ дают противоположные углы: $\\angle A+\\angle C=180°$ и $\\angle B+\\angle D=180°$. Углы $A$ и $B$ — соседние, их сумма не обязательно равна $180°$. +
Ответ: в)
` + }, + { + text: `Найдите значение выражения $10A$, + если $A = \\sqrt{0{,}36} \\cdot \\sqrt{100} - \\sqrt{1{,}69}$.`, + sol: `Извлекаем корни: +$$\\sqrt{0{,}36}=0{,}6;\\quad \\sqrt{100}=10;\\quad \\sqrt{1{,}69}=1{,}3$$ +Подставляем: +$$A = 0{,}6\\cdot 10 - 1{,}3 = 6 - 1{,}3 = 4{,}7$$ +$$10A = 10\\cdot 4{,}7 = 47$$ +
Ответ: $47$
` + }, + { + text: `Первый член арифметической прогрессии равен $-3\\dfrac{3}{4}$, + разность прогрессии равна $-0{,}25$. + Является ли членом данной прогрессии число $-8$? Ответ обоснуйте.`, + sol: `Дано: $a_1 = -3\\dfrac{3}{4} = -\\dfrac{15}{4}$, $d = -0{,}25 = -\\dfrac{1}{4}$. +
Формула $n$-го члена: $a_n = a_1 + (n-1)d$. +
Если $-8$ — член прогрессии, найдём его номер $n$: +$$-8 = -\\dfrac{15}{4} + (n-1)\\cdot\\left(-\\dfrac{1}{4}\\right)$$ +Умножим обе части на $-4$: +$$32 = 15 + (n-1)$$ +$$n-1 = 17 \\implies n = 18$$ +Получили натуральное число $n=18$, значит $-8$ — это 18-й член прогрессии. +
Ответ: да, $-8$ — член прогрессии (18-й по счёту)
` + }, + { + text: `$ABCD$ — прямоугольник с периметром $22$ см, + сторона $AB$ на $5$ см меньше стороны $AD$. + Найдите площадь прямоугольника.`, + sol: `Формула периметра прямоугольника: $P = 2(a+b)$, где $a$ и $b$ — соседние стороны. +
Формула площади прямоугольника: $S = a\\cdot b$. +
Шаг 1. Обозначим $AD = x$ см. Тогда по условию $AB = x-5$ см, так как сторона $AB$ на $5$ см меньше $AD$. +
Шаг 2. Составим уравнение по формуле периметра: +$$2(AD+AB) = 22 \\;\\implies\\; 2(x + x-5) = 22$$ +$$2(2x-5) = 22 \\;\\implies\\; 2x-5 = 11 \\;\\implies\\; x = 8$$ +Шаг 3. Находим стороны: $AD = 8$ см, $AB = 8-5 = 3$ см. + + + A + D + C + B + AD = 8 + AB=3 + +Шаг 4. Применяем формулу площади: +$$S = AD\\cdot AB = 8\\cdot 3 = 24\\text{ см}^2$$ +
Ответ: $24$ см²
` + }, + { + text: `Оптовая стоимость справочного издания «Памятные места Беларуси» $20$ р. + Какое наибольшее количество данных книг по розничной цене можно купить на $7000$ р., + если розничная цена на $30\\%$ выше оптовой?`, + sol: `Правило увеличения числа на $p$ процентов: новое значение равно $N\\cdot\\left(1+\\dfrac{p}{100}\\right)$. +
Шаг 1. Найдём розничную цену. По условию она на $30\\%$ выше оптовой, значит: +$$20\\cdot\\left(1+\\dfrac{30}{100}\\right) = 20\\cdot 1{,}3 = 26\\text{ р.}$$ +Шаг 2. Чтобы узнать, сколько книг можно купить на $7000$ р., делим бюджет на цену одной книги: +$$\\dfrac{7000}{26} = 269{,}23\\ldots$$ +Шаг 3. Количество книг — натуральное число, поэтому округляем результат вниз: получаем $269$ книг. +
Шаг 4. Проверим граничные значения: +
$\\bullet$ $269\\cdot 26 = 6994$ р. — меньше $7000$, значит на $269$ книг денег хватит; +
$\\bullet$ $270\\cdot 26 = 7020$ р. — больше $7000$, значит $270$ книг купить уже нельзя. +
Ответ: $269$ книг
` + }, + { + text: `Упростите выражение + $$\\left(\\dfrac{x-4}{x^2-2x+1} - \\dfrac{x+2}{x^2+x-2}\\right) : \\dfrac{1}{(2x-2)^2}.$$`, + sol: `Формула квадрата разности: $a^2-2ab+b^2=(a-b)^2$. Разложение квадратного трёхчлена: $x^2+x-2=(x+2)(x-1)$, так как корни $-2$ и $1$. Правило деления дробей: $\\dfrac{A}{B}:\\dfrac{C}{D}=\\dfrac{A}{B}\\cdot\\dfrac{D}{C}$. +
Шаг 1. Разложим все знаменатели на множители: +$$x^2-2x+1 = (x-1)^2;$$ +$$x^2+x-2 = (x+2)(x-1);$$ +$$(2x-2)^2 = \\bigl(2(x-1)\\bigr)^2 = 4(x-1)^2.$$ +Шаг 2. Найдём ОДЗ: знаменатели не равны нулю, значит $x\\neq 1$ и $x\\neq -2$. +
Шаг 3. Сократим вторую дробь в скобках — множитель $(x+2)$ есть в числителе и в знаменателе: +$$\\dfrac{x+2}{(x+2)(x-1)} = \\dfrac{1}{x-1}.$$ +Шаг 4. Приведём дроби в скобках к общему знаменателю $(x-1)^2$. Дробь $\\dfrac{1}{x-1}$ умножаем на $\\dfrac{x-1}{x-1}$: +$$\\dfrac{x-4}{(x-1)^2} - \\dfrac{x-1}{(x-1)^2} = \\dfrac{(x-4)-(x-1)}{(x-1)^2} = \\dfrac{-3}{(x-1)^2}.$$ +Шаг 5. По правилу деления заменяем деление умножением на обратную дробь: +$$\\dfrac{-3}{(x-1)^2} : \\dfrac{1}{(2x-2)^2} = \\dfrac{-3}{(x-1)^2}\\cdot 4(x-1)^2.$$ +Множители $(x-1)^2$ сокращаются: +$$\\dfrac{-3}{(x-1)^2}\\cdot 4(x-1)^2 = -3\\cdot 4 = -12.$$ +
Ответ: $-12$
` + }, + { + text: `Найдите сумму координат точки пересечения прямых, + заданных уравнениями $1{,}2x + 3{,}4y = 12$ и $2{,}5x + 1{,}4y = 25$.`, + sol: `Метод решения: чтобы найти точку пересечения двух прямых, решаем систему из их уравнений. Используем метод сложения (исключаем одну переменную). +
Шаг 1. Запишем систему: +$$\\begin{cases} 1{,}2x + 3{,}4y = 12 \\\\ 2{,}5x + 1{,}4y = 25 \\end{cases}$$ +Шаг 2. Уравняем коэффициенты при $x$. Умножим первое уравнение на $25$, второе — на $12$: +$$\\begin{cases} 30x + 85y = 300 \\\\ 30x + 16{,}8y = 300 \\end{cases}$$ +Шаг 3. Вычтем второе уравнение из первого: +$$(85 - 16{,}8)y = 0 \\;\\implies\\; 68{,}2y = 0 \\;\\implies\\; y = 0$$ +Шаг 4. Подставим $y=0$ в первое уравнение системы: +$$1{,}2x = 12 \\;\\implies\\; x = 10$$ +Шаг 5. Значит, точка пересечения — $(10;\\,0)$. Тогда сумма координат: +$$x+y = 10+0 = 10$$ +
Ответ: $10$
` + }, + { + text: `Диагонали трапеции $ABCD$ ($AD \\| BC$) пересекаются в точке $O$. + Площади треугольников $ABO$ и $BOC$ равны соответственно $16$ см² и $8$ см². + Найдите площадь трапеции.`, + sol: ` + + + + + + + + + O + A + D + B + C + 16 + 8 + ? + ? + +Шаг 1 — отношение оснований.
+Треугольники $ABO$ и $BOC$ имеют общую вершину $B$, а основания $AO$ и $OC$ лежат на одной прямой $AC$. Значит, отношение площадей равно отношению оснований: +$$\\dfrac{S_{ABO}}{S_{BOC}} = \\dfrac{AO}{OC} = \\dfrac{16}{8} = 2$$ +По свойству трапеции $\\dfrac{AO}{OC} = \\dfrac{AD}{BC}$, значит $\\dfrac{AD}{BC} = 2$. +

+Шаг 2 — площадь треугольника AOD.
+Треугольники $AOD$ и $BOC$ подобны (т.к. $AD\\parallel BC$), коэффициент подобия $k = \\dfrac{AD}{BC} = 2$. Отношение площадей $= k^2 = 4$: +$$S_{AOD} = 4\\cdot S_{BOC} = 4\\cdot 8 = 32\\text{ см}^2$$ +Шаг 3 — площадь треугольника COD.
+В трапеции с диагоналями: $S_{ABO} = S_{COD}$ (известное свойство). Значит, $S_{COD} = 16$ см². +

+Шаг 4 — итог. +$$S_{ABCD} = S_{ABO}+S_{BOC}+S_{COD}+S_{AOD} = 16+8+16+32 = 72\\text{ см}^2$$ +
Ответ: $72$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v20.js b/frontend/js/exam9/variants/v20.js new file mode 100644 index 0000000..3592317 --- /dev/null +++ b/frontend/js/exam9/variants/v20.js @@ -0,0 +1,198 @@ +VARIANTS[20] = { + label: "Вариант 20", + tasks: [ + { + text: `Определите рисунок, на котором изображён график функции $y = (x-1)^2$:`, + figure: ``, + sol: `Функция $y=(x-1)^2$ — это парабола $y=x^2$, сдвинутая на $1$ единицу вправо. +
    +
  • Вершина параболы: $(1;\\,0)$ — точка на оси $Ox$
    • +
  • $y$-пересечение: при $x=0$, $y=1$ — точка $(0;\\,1)$
    • +
  • Ветви параболы направлены вверх
    • +
  • График симметричен относительно прямой $x=1$
    • +
+На рисунке ищем параболу с вершиной в точке $(1;\\,0)$, ветви направлены вверх. +
Ответ: рисунок с параболой, вершина которой в точке $(1;\\,0)$
` + }, + { + text: `Какие из данных чисел являются решениями системы неравенств + $$\\left\\{\\begin{array}{l} a > -4, \\\\[4pt] a \\leq 3\\dfrac{1}{2}; \\end{array}\\right.$$`, + opts: [ + ["а", "$3$"], ["б", "$-4$"], ["в", "$0$"], + ["г", "$4{,}5$"], ["д", "$-4{,}5$"], + ], + sol: `Решение системы: $-4 < a \\leq 3{,}5$. + + + + −4 + 0 + 3,5 + + + + +Проверяем каждое: +
    +
  • а) $3$: $-4 < 3 \\leq 3{,}5$ ✓
    • +
  • б) $-4$: условие $a>-4$ нарушено ✗
    • +
  • в) $0$: $-4 < 0 \\leq 3{,}5$ ✓
    • +
  • г) $4{,}5 > 3{,}5$ ✗
    • +
  • д) $-4{,}5 < -4$ ✗
    • +
+
Ответ: а) и в)
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "диагонали равнобедренной трапеции точкой пересечения делятся пополам;"], + ["б", "периметр квадрата со стороной $6$ см равен $24$ см;"], + ["в", "если у четырёхугольника $ABCD$ $\\angle A + \\angle C = 180^{\\circ}$, то около него можно описать окружность;"], + ["г", "если у $\\triangle ABC$ $\\angle C = 90^{\\circ}$, $AC = 3$, $BC = 4$, то $\\operatorname{ctg} A = \\dfrac{3}{4}$?"], + ], + sol: `
    +
  • а) Диагонали равнобедренной трапеции равны, но делятся точкой пересечения не пополам — НЕВЕРНО
    • +
  • б) Периметр квадрата со стороной $6$: $4\\cdot6=24$ см — верно
    • +
  • в) Условие вписанности четырёхугольника: $\\angle A+\\angle C=180°$ — верно
    • +
  • г) $\\operatorname{ctg}A = \\dfrac{AC}{BC}=\\dfrac{3}{4}$ — верно
    • +
+Диагонали равнобедренной трапеции равны по длине, но не делятся пополам в точке пересечения (это свойство параллелограмма). +
Ответ: а)
` + }, + { + text: `Найдите значение выражения $10A$, + если $A = \\sqrt{0{,}49} \\cdot \\sqrt{25} - \\sqrt{1{,}96}$.`, + sol: `Извлекаем корни: +$$\\sqrt{0{,}49}=0{,}7;\\quad \\sqrt{25}=5;\\quad \\sqrt{1{,}96}=1{,}4$$ +Подставляем: +$$A = 0{,}7\\cdot 5 - 1{,}4 = 3{,}5 - 1{,}4 = 2{,}1$$ +$$10A = 10\\cdot 2{,}1 = 21$$ +
Ответ: $21$
` + }, + { + text: `Первый член арифметической прогрессии равен $-4\\dfrac{1}{2}$, + разность прогрессии равна $-0{,}5$. + Является ли членом данной прогрессии число $-10$? Ответ обоснуйте.`, + sol: `Дано: $a_1 = -4\\dfrac{1}{2} = -\\dfrac{9}{2}$, $d = -0{,}5 = -\\dfrac{1}{2}$. +
Формула $n$-го члена: $a_n = a_1 + (n-1)d$. +
Если $-10$ — член прогрессии, найдём его номер $n$: +$$-10 = -\\dfrac{9}{2} + (n-1)\\cdot\\left(-\\dfrac{1}{2}\\right)$$ +Умножим обе части на $-2$: +$$20 = 9 + (n-1)$$ +$$n-1 = 11 \\implies n = 12$$ +Получили натуральное число $n=12$, значит $-10$ — это 12-й член прогрессии. +
Ответ: да, $-10$ — член прогрессии (12-й по счёту)
` + }, + { + text: `$ABCD$ — прямоугольник с периметром $36$ см, + сторона $AD$ в $2$ раза больше стороны $AB$. + Найдите площадь прямоугольника.`, + sol: `Формула периметра прямоугольника: $P = 2(a+b)$. +
Формула площади прямоугольника: $S = a\\cdot b$. +
Шаг 1. Введём обозначение: пусть $AB = x$ см. Тогда $AD = 2x$ см, так как по условию $AD$ в $2$ раза больше $AB$. +
Шаг 2. Составим уравнение по формуле периметра: +$$2(AB+AD) = 36 \\;\\implies\\; 2(x+2x) = 36 \\;\\implies\\; 6x = 36 \\;\\implies\\; x = 6$$ +Шаг 3. Находим стороны: $AB = 6$ см, $AD = 12$ см. + + + A + D + C + B + AD = 12 + AB=6 + +Шаг 4. Подставляем в формулу площади: +$$S = AD\\cdot AB = 12\\cdot 6 = 72\\text{ см}^2$$ +
Ответ: $72$ см²
` + }, + { + text: `Оптовая стоимость справочного издания «Образование в Беларуси: истоки, история, современность» $28$ р. + Какое наибольшее количество данных книг по розничной цене можно купить на $5500$ р., + если розничная цена на $25\\%$ выше оптовой?`, + sol: `Правило увеличения числа на $p$ процентов: новое значение равно $N\\cdot\\left(1+\\dfrac{p}{100}\\right)$. +
Шаг 1. Найдём розничную цену. Так как она на $25\\%$ выше оптовой: +$$28\\cdot\\left(1+\\dfrac{25}{100}\\right) = 28\\cdot 1{,}25 = 35\\text{ р.}$$ +Шаг 2. Делим имеющуюся сумму на цену одной книги, чтобы узнать, сколько книг можно купить: +$$\\dfrac{5500}{35} = 157{,}14\\ldots$$ +Шаг 3. Так как количество книг — натуральное число, округляем результат вниз: получаем $157$ книг. +
Шаг 4. Проверим граничные значения: +
$\\bullet$ $157\\cdot 35 = 5495$ р. — меньше $5500$, значит $157$ книг купить можно; +
$\\bullet$ $158\\cdot 35 = 5530$ р. — больше $5500$, значит $158$ книг купить уже нельзя. +
Ответ: $157$ книг
` + }, + { + text: `Упростите выражение + $$\\left(\\dfrac{y+2}{y^2-y-6} - \\dfrac{y}{y^2-6y+9}\\right) : \\dfrac{1}{(2y-6)^2}.$$`, + sol: `Формула квадрата разности: $a^2-2ab+b^2=(a-b)^2$. Разложение квадратного трёхчлена: $y^2-y-6=(y-3)(y+2)$, так как корни $3$ и $-2$. Правило деления дробей: $\\dfrac{A}{B}:\\dfrac{C}{D}=\\dfrac{A}{B}\\cdot\\dfrac{D}{C}$. +
Шаг 1. Разложим знаменатели на множители: +$$y^2-y-6 = (y-3)(y+2);$$ +$$y^2-6y+9 = (y-3)^2;$$ +$$(2y-6)^2 = \\bigl(2(y-3)\\bigr)^2 = 4(y-3)^2.$$ +Шаг 2. Запишем ОДЗ: знаменатели не равны нулю, значит $y\\neq 3$ и $y\\neq -2$. +
Шаг 3. Сократим первую дробь в скобках — множитель $(y+2)$ есть в числителе и в знаменателе: +$$\\dfrac{y+2}{(y-3)(y+2)} = \\dfrac{1}{y-3}.$$ +Шаг 4. Приведём дроби в скобках к общему знаменателю $(y-3)^2$. Дробь $\\dfrac{1}{y-3}$ умножим на $\\dfrac{y-3}{y-3}$: +$$\\dfrac{y-3}{(y-3)^2} - \\dfrac{y}{(y-3)^2} = \\dfrac{(y-3)-y}{(y-3)^2} = \\dfrac{-3}{(y-3)^2}.$$ +Шаг 5. Заменим деление умножением на обратную дробь: +$$\\dfrac{-3}{(y-3)^2} : \\dfrac{1}{(2y-6)^2} = \\dfrac{-3}{(y-3)^2}\\cdot 4(y-3)^2.$$ +Сокращаем $(y-3)^2$: +$$\\dfrac{-3}{(y-3)^2}\\cdot 4(y-3)^2 = -3\\cdot 4 = -12.$$ +
Ответ: $-12$
` + }, + { + text: `Найдите сумму координат точки пересечения прямых, + заданных уравнениями $2x + 3y = -23$ и $x - y = 11(10 + y)$.`, + sol: `Метод решения: точка пересечения двух прямых — общее решение их уравнений. Используем метод подстановки. +
Шаг 1. Упростим второе уравнение, раскрыв скобки в правой части: +$$x - y = 110 + 11y \\;\\implies\\; x = 110 + 12y$$ +Шаг 2. Подставим $x = 110+12y$ в первое уравнение и решим относительно $y$: +$$2(110+12y)+3y = -23$$ +$$220+24y+3y = -23$$ +$$27y = -243 \\;\\implies\\; y = -9$$ +Шаг 3. Найдём $x$, подставив $y=-9$: +$$x = 110 + 12\\cdot(-9) = 110 - 108 = 2$$ +Шаг 4. Точка пересечения — $(2;\\,-9)$. Сумма координат: +$$x+y = 2+(-9) = -7$$ +
Ответ: $-7$
` + }, + { + text: `Диагонали трапеции $ABCD$ ($AD \\| BC$) пересекаются в точке $O$. + Площади треугольников $AOD$ и $COD$ равны соответственно $54$ см² и $18$ см². + Найдите площадь трапеции.`, + sol: ` + + + + + + + + + O + A + D + B + C + 18 + 6 + 54 + 18 + +Шаг 1 — отношение оснований.
+Треугольники $AOD$ и $COD$ имеют общую вершину $D$, основания $AO$ и $OC$ лежат на диагонали $AC$. Отношение площадей равно отношению оснований: +$$\\dfrac{S_{AOD}}{S_{COD}} = \\dfrac{AO}{OC} = \\dfrac{54}{18} = 3$$ +По свойству трапеции $\\dfrac{AO}{OC} = \\dfrac{AD}{BC}$, значит $\\dfrac{AD}{BC} = 3$. +

+Шаг 2 — площадь треугольника BOC.
+Треугольники $AOD$ и $BOC$ подобны (т.к. $AD\\parallel BC$), коэффициент подобия $k = 3$. Отношение площадей $= k^2 = 9$: +$$S_{BOC} = \\dfrac{S_{AOD}}{9} = \\dfrac{54}{9} = 6\\text{ см}^2$$ +Шаг 3 — площадь треугольника ABO.
+В трапеции: $S_{ABO} = S_{COD}$ (известное свойство). Значит, $S_{ABO} = 18$ см². +

+Шаг 4 — итог. +$$S_{ABCD} = S_{ABO}+S_{BOC}+S_{COD}+S_{AOD} = 18+6+18+54 = 96\\text{ см}^2$$ +
Ответ: $96$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v21.js b/frontend/js/exam9/variants/v21.js new file mode 100644 index 0000000..bdc1bb4 --- /dev/null +++ b/frontend/js/exam9/variants/v21.js @@ -0,0 +1,212 @@ +VARIANTS[21] = { + label: "Вариант 21", + tasks: [ + { + text: `Значение выражения $\\dfrac{3}{4} : \\dfrac{7}{3}$ равно:`, + opts: [ + ["а", "$\\dfrac{28}{9}$"], ["б", "$\\dfrac{21}{12}$"], ["в", "$\\dfrac{9}{28}$"], + ["г", "$\\dfrac{7}{4}$"], ["д", "$\\dfrac{4}{7}$"], + ], + sol: `Деление дробей — умножаем первую на обратную второй: +$$\\dfrac{3}{4} : \\dfrac{7}{3} = \\dfrac{3}{4} \\cdot \\dfrac{3}{7} = \\dfrac{9}{28}$$ +
Ответ: в) $\\dfrac{9}{28}$
` + }, + { + text: `Определите, какое из чисел является решением уравнения $3 = -6x$:`, + opts: [ + ["а", "$-2$"], ["б", "$2$"], ["в", "$\\dfrac{1}{2}$"], + ["г", "$-\\dfrac{1}{2}$"], ["д", "$-3$"], + ], + sol: `$$3 = -6x \\implies x = \\dfrac{3}{-6} = -\\dfrac{1}{2}$$ +Проверка: $-6\\cdot\\bigl(-\\tfrac{1}{2}\\bigr)=3$ ✓ +
Ответ: г) $-\\dfrac{1}{2}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "сумма смежных углов равна $180^{\\circ}$;"], + ["б", "на плоскости две прямые, параллельные третьей, параллельны между собой;"], + ["в", "в любом параллелограмме противоположные углы равны между собой;"], + ["г", "существует треугольник со сторонами, равными $3$ см, $5$ см и $8$ см?"], + ], + sol: `
    +
  • а) Сумма смежных углов $=180°$ — верно
    • +
  • б) Транзитивность параллельности — верно
    • +
  • в) Противоположные углы параллелограмма равны — верно
    • +
  • г) Стороны $3$, $5$, $8$: $3+5=8$, но для треугольника нужно строгое неравенство $a+b>c$ — НЕВЕРНО (вырожденный «треугольник»)
    • +
+
Ответ: г)
` + }, + { + text: `Упростите выражение $2(a - 3b) - 3(2a + b)$.`, + sol: `Раскрываем скобки и собираем подобные: +$$2(a-3b)-3(2a+b) = 2a-6b-6a-3b = -4a-9b$$ +
Ответ: $-4a-9b$
` + }, + { + text: `Таня затратила на выполнение домашнего задания $2$ ч. + На выполнение домашнего задания по алгебре она затратила $\\dfrac{1}{3}$ всего времени. + Сколько минут Таня выполняла оставшуюся часть домашнего задания?`, + sol: `Правило нахождения части от числа: чтобы найти $\\dfrac{m}{n}$ от числа $A$, умножают $A$ на $\\dfrac{m}{n}$. +
Шаг 1. Переведём общее время в минуты, так как ответ нужен именно в минутах: +$$2\\text{ ч} = 2\\cdot 60 = 120\\text{ мин}.$$ +Шаг 2. Найдём время, затраченное на алгебру. По условию это $\\dfrac{1}{3}$ всего времени: +$$\\dfrac{1}{3}\\cdot 120 = 40\\text{ мин}.$$ +Шаг 3. Оставшаяся часть задания — это разность общего времени и времени на алгебру: +$$120 - 40 = 80\\text{ мин}.$$ +
Ответ: $80$ мин
` + }, + { + text: `Сумма длин гипотенузы и катета, лежащего в данном треугольнике против угла в $30^{\\circ}$, + равна $24$ см. Найдите площадь круга, описанного около треугольника.`, + sol: `Свойство прямоугольного треугольника с углом $30°$: катет, лежащий против угла $30°$, равен половине гипотенузы. +
Свойство описанной окружности прямоугольного треугольника: центр окружности лежит в середине гипотенузы, поэтому $R = \\dfrac{c}{2}$, где $c$ — гипотенуза. +
Формула площади круга: $S = \\pi R^2$. +
Шаг 1. Обозначим гипотенузу через $c$. Тогда катет, лежащий против угла $30°$, равен $\\dfrac{c}{2}$. +
Шаг 2. Составим уравнение из условия «сумма равна $24$»: +$$c + \\dfrac{c}{2} = 24 \\;\\implies\\; \\dfrac{3c}{2} = 24 \\;\\implies\\; c = 16\\text{ см}$$ + + + + + + + + A + B + C + O + R + 30° + AB = 16 см + BC=8 + +Шаг 3. Так как гипотенуза — диаметр описанной окружности, радиус: +$$R = \\dfrac{c}{2} = \\dfrac{16}{2} = 8\\text{ см}$$ +Шаг 4. Применяем формулу площади круга: +$$S = \\pi R^2 = \\pi\\cdot 8^2 = 64\\pi\\text{ см}^2$$ +
Ответ: $64\\pi$ см²
` + }, + { + text: `Найдите наибольшее целое решение неравенства + $\\dfrac{(x-9)^2}{x^2+x-12} \\leq 0$.`, + sol: `Метод интервалов для дробного неравенства: сначала находим ОДЗ (знаменатель $\\neq 0$), затем определяем знаки числителя и знаменателя. +
Шаг 1. Разложим знаменатель на множители. Ищем числа с произведением $-12$ и суммой $1$ — это $4$ и $-3$: +$$x^2+x-12=(x+4)(x-3)$$ +ОДЗ: $x\\neq -4$ и $x\\neq 3$, иначе знаменатель равен нулю. +
Шаг 2. Числитель $(x-9)^2$ — квадрат, значит он всегда неотрицателен. Дробь будет $\\leq 0$ в двух случаях: +
    +
  • числитель равен нулю, то есть $x=9$ (тогда вся дробь $=0$; знаменатель при $x=9$ равен $9^2+9-12=78\\neq 0$, ОДЗ соблюдено);
    • +
  • числитель строго больше нуля, а знаменатель отрицателен (дробь отрицательна).
    • +
+Шаг 3. Решаем неравенство $(x+4)(x-3)\\lt 0$. Произведение двух множителей отрицательно, когда они разных знаков, поэтому $-4\\lt x\\lt 3$. +
Шаг 4. Объединяем оба случая: $x\\in(-4;\\,3)\\cup\\{9\\}$. +
Шаг 5. Целые числа в $(-4;\\;3)$: $-3,-2,-1,0,1,2$. И ещё $x=9$. +
Наибольшее из всех целых решений — $x=9$. +
Ответ: наибольшее целое решение $x = 9$
` + }, + { + text: `Постройте графики уравнений системы + $$\\begin{cases} x + y = 5, \\\\[4pt] y - x^2 = 3 \\end{cases}$$ + и найдите сумму ординат точек пересечения.`, + sol: `Метод решения: точки пересечения графиков — это решения системы. Решаем методом подстановки. +
Шаг 1. Из первого уравнения выразим $y$: $y = 5 - x$. +
Шаг 2. Подставим во второе уравнение $y - x^2 = 3$: +$$(5-x)-x^2 = 3 \\;\\implies\\; -x^2 - x + 2 = 0 \\;\\implies\\; x^2+x-2=0$$ +Шаг 3. Раскладываем на множители (по теореме Виета: сумма корней $-1$, произведение $-2$, значит корни $-2$ и $1$): +$$(x+2)(x-1)=0 \\;\\implies\\; x_1 = -2,\\; x_2 = 1$$ +Шаг 4. Находим $y$ по формуле $y = 5 - x$: +
при $x = -2$: $y = 5 - (-2) = 7$; +
при $x = 1$: $y = 5 - 1 = 4$. + + + + + + + + −2 + −1 + 1 + 2 + 3 + 7 + 4 + 3 + + + + + (−2; 7) + (1; 4) + y=5−x + y=x²+3 + x + y + +Шаг 5. Точки пересечения: $(-2;\\,7)$ и $(1;\\,4)$. Сумма ординат: +$$7 + 4 = 11$$ +
Ответ: $11$
` + }, + { + text: `Три числа, дающие в сумме $36$, являются последовательными членами арифметической прогрессии. + Если от первого числа вычесть $5$, от второго вычесть $6$, а третье число увеличить вдвое, + то полученные числа будут последовательными членами геометрической прогрессии. + Найдите эти числа.`, + sol: `Пусть три члена АП: $12-d$, $12$, $12+d$ (сумма $=3\\cdot12=36$). +
После преобразований получаем три члена ГП: $(7-d)$, $6$, $(24+2d)$. +
Условие ГП ($b^2=ac$): +$$6^2=(7-d)(24+2d) \\implies 36=168-10d-2d^2$$ +$$2d^2+10d-132=0 \\implies d^2+5d-66=0$$ +$$D=25+264=289=17^2 \\implies d=\\dfrac{-5\\pm17}{2}$$ +$d=6$: АП: $6,12,18$. ГП: $1,6,36$ (знаменатель $6$) ✓ +
$d=-11$: АП: $23,12,1$. ГП: $18,6,2$ (знаменатель $\\tfrac{1}{3}$) ✓ +
Ответ: $6,\\ 12,\\ 18$ или $1,\\ 12,\\ 23$
` + }, + { + text: `$AM$ — медиана треугольника $ABC$, площадь которого $120$ см². + Точка $E$ — середина медианы $AM$. + Луч $BE$ пересекает сторону $AC$ в точке $K$. + Найдите площадь четырёхугольника $МЕКС$.`, + sol: ` + + + + + + + + + A + B + C + M + E + K + МЕКС + +Шаг 1. Медиана $AM$ делит $\\triangle ABC$ пополам: +$$S_{ABM} = S_{ACM} = 60\\text{ см}^2$$ +Шаг 2. Ключевой факт: $S_{ABK} = S_{MBK}$. +
Треугольники $ABK$ и $MBK$ имеют общее основание $BK$. Точка $E$ — середина $AM$ — лежит на луче $BK$ (по условию). Значит, прямая $BK$ пересекает отрезок $AM$ ровно в его середине, то есть $A$ и $M$ находятся по разные стороны от прямой $BK$ на одинаковом расстоянии. Следовательно, высоты из $A$ и $M$ на $BK$ равны, и: +$$S_{ABK} = S_{MBK}$$ +Шаг 3. $M$ — середина $BC$ ⟹ треугольники $MBK$ и $CBK$ имеют основания $MB$ и $CB$ при одинаковой высоте из $K$: +$$\\dfrac{S_{MBK}}{S_{CBK}} = \\dfrac{MB}{CB} = \\dfrac{1}{2} \\implies S_{CBK} = 2\\cdot S_{ABK}$$ +Шаг 4. Обозначим $S_{ABK} = p$. Тогда $S_{CBK} = 2p$. Точка $K$ на $AC$ делит $\\triangle ABC$ на два треугольника: +$$p + 2p = 120 \\implies p = 40$$ +Итак, $S_{ABK}=40$ см², $S_{CBK}=80$ см². +
Шаг 5. $M$ — середина $BC$ ⟹ +$$S_{MKC} = \\dfrac{1}{2}\\cdot S_{CBK} = 40\\text{ см}^2$$ +Шаг 6. Из $S_{ABK}=40$ находим $AK:KC$. Треугольники $ABK$ и $ABC$ — общая вершина $B$, основания $AK$ и $AC$: +$$\\dfrac{AK}{AC} = \\dfrac{S_{ABK}}{S_{ABC}} = \\dfrac{40}{120} = \\dfrac{1}{3}$$ +Шаг 7. Треугольники $AMK$ и $AMC$ — общая вершина $M$, основания на $AC$: +$$S_{AMK} = \\dfrac{AK}{AC}\\cdot S_{AMC} = \\dfrac{1}{3}\\cdot 60 = 20\\text{ см}^2$$ +Шаг 8. $E$ — середина $AM$ ⟹ треугольники $AEK$ и $AMK$ имеют основания $AE$ и $AM$: +$$S_{AEK} = \\dfrac{AE}{AM}\\cdot S_{AMK} = \\dfrac{1}{2}\\cdot 20 = 10\\text{ см}^2$$ +Шаг 9. $S_{MEK} = S_{AMK} - S_{AEK} = 20 - 10 = 10$ см². +
Итог: +$$S_{МЕКС} = S_{MEK} + S_{MKC} = 10 + 40 = 50\\text{ см}^2$$ +
Ответ: $50$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v22.js b/frontend/js/exam9/variants/v22.js new file mode 100644 index 0000000..65051e6 --- /dev/null +++ b/frontend/js/exam9/variants/v22.js @@ -0,0 +1,202 @@ +VARIANTS[22] = { + label: "Вариант 22", + tasks: [ + { + text: `Значение выражения $\\dfrac{3}{7} : \\dfrac{7}{8}$ равно:`, + opts: [ + ["а", "$\\dfrac{49}{24}$"], ["б", "$\\dfrac{21}{8}$"], ["в", "$\\dfrac{24}{49}$"], + ["г", "$\\dfrac{3}{8}$"], ["д", "$\\dfrac{8}{3}$"], + ], + sol: `Деление дробей — умножаем первую на обратную второй: +$$\\dfrac{3}{7} : \\dfrac{7}{8} = \\dfrac{3}{7} \\cdot \\dfrac{8}{7} = \\dfrac{24}{49}$$ +
Ответ: в) $\\dfrac{24}{49}$
` + }, + { + text: `Определите, какое из чисел является решением уравнения $-5 = 10x$:`, + opts: [ + ["а", "$-2$"], ["б", "$2$"], ["в", "$\\dfrac{1}{2}$"], + ["г", "$-\\dfrac{1}{2}$"], ["д", "$-5$"], + ], + sol: `$$-5 = 10x \\implies x = \\dfrac{-5}{10} = -\\dfrac{1}{2}$$ +Проверка: $10\\cdot\\bigl(-\\tfrac{1}{2}\\bigr)=-5$ ✓ +
Ответ: г) $-\\dfrac{1}{2}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "вертикальные углы равны;"], + ["б", "на плоскости две прямые, перпендикулярные третьей, параллельны между собой;"], + ["в", "в любой трапеции углы, прилежащие к боковой стороне, в сумме равны $180^{\\circ}$;"], + ["г", "существует треугольник со сторонами, равными $10$ см, $6$ см и $4$ см?"], + ], + sol: `
    +
  • а) Вертикальные углы равны — верно
    • +
  • б) На плоскости две прямые, перпендикулярные третьей, параллельны между собой — верно
    • +
  • в) В любой трапеции углы, прилежащие к боковой стороне, в сумме равны $180^{\\circ}$ — верно
    • +
  • г) Стороны $10$, $6$, $4$: $6+4=10$, но для треугольника нужно строгое неравенство $a+b>c$ — НЕВЕРНО (вырожденный «треугольник»)
    • +
+
Ответ: г)
` + }, + { + text: `Упростите выражение $-2(m - n) - 3(m + n)$.`, + sol: `Раскрываем скобки и собираем подобные: +$$-2(m-n)-3(m+n) = -2m+2n-3m-3n = -5m-n$$ +
Ответ: $-5m-n$
` + }, + { + text: `На уроке математики, который длится $45$ минут, $\\dfrac{4}{9}$ всего времени + учащиеся выполняли самостоятельную работу, а оставшееся время изучали новую тему. + Сколько минут учащиеся изучали новую тему?`, + sol: `Правило нахождения части от числа: чтобы найти $\\dfrac{m}{n}$ от числа $A$, умножают $A$ на $\\dfrac{m}{n}$. +
Шаг 1. Найдём время, затраченное на самостоятельную работу. По условию оно составляет $\\dfrac{4}{9}$ от всего времени урока: +$$\\dfrac{4}{9}\\cdot 45 = \\dfrac{4\\cdot 45}{9} = \\dfrac{180}{9} = 20\\text{ мин}.$$ +Шаг 2. Оставшееся время — это разность между длительностью урока и временем на самостоятельную работу: +$$45 - 20 = 25\\text{ мин}.$$ +Значит, новую тему изучали $25$ минут. +
Ответ: $25$ минут
` + }, + { + text: `Разность длин гипотенузы и катета, лежащего в данном треугольнике против угла в $30^{\\circ}$, + равна $12$ см. Найдите длину окружности, описанной около треугольника.`, + sol: `Свойство прямоугольного треугольника с углом $30°$: катет, лежащий против угла $30°$, равен половине гипотенузы. +
Свойство описанной окружности прямоугольного треугольника: гипотенуза является диаметром окружности, описанной около треугольника, поэтому $R=\\dfrac{c}{2}$. +
Формула длины окружности: $L = 2\\pi R$. +
Шаг 1. Обозначим гипотенузу $c$. Тогда катет против $30°$ равен $\\dfrac{c}{2}$. +
Шаг 2. Составим уравнение из условия «разность равна $12$»: +$$c - \\dfrac{c}{2} = 12 \\;\\implies\\; \\dfrac{c}{2} = 12 \\;\\implies\\; c = 24\\text{ см}$$ + + + + + + + + A + B + C + O + R + 30° + AB = 24 см + BC=12 + +Шаг 3. Радиус описанной окружности: +$$R = \\dfrac{c}{2} = \\dfrac{24}{2} = 12\\text{ см}$$ +Шаг 4. Применяем формулу длины окружности: +$$L = 2\\pi R = 2\\pi\\cdot 12 = 24\\pi\\text{ см}$$ +
Ответ: $24\\pi$ см
` + }, + { + text: `Найдите наименьшее целое решение неравенства + $\\dfrac{(x-5)^2}{x^2+x-20} \\leq 0$.`, + sol: `Метод интервалов для дробного неравенства: сначала находим ОДЗ (знаменатель $\\neq 0$), затем определяем знаки числителя и знаменателя. +
Шаг 1. Разложим знаменатель на множители. Ищем два числа с произведением $-20$ и суммой $1$ — это $5$ и $-4$: +$$x^2+x-20=(x+5)(x-4)$$ +ОДЗ: $x\\neq -5$ и $x\\neq 4$, иначе знаменатель равен нулю. +
Шаг 2. Числитель $(x-5)^2$ — квадрат, поэтому $\\geq 0$. Значит дробь $\\leq 0$ в двух случаях: +
    +
  • числитель равен нулю: $x=5$ (знаменатель при $x=5$ равен $25+5-20=10\\neq 0$, ОДЗ соблюдено);
    • +
  • числитель строго положителен, а знаменатель отрицателен.
    • +
+Шаг 3. Решаем $(x+5)(x-4)\\lt 0$. Произведение отрицательно, когда множители разных знаков: $-5\\lt x\\lt 4$. +
Шаг 4. Объединяем: $x\\in(-5;\\,4)\\cup\\{5\\}$. +
Шаг 5. Целые числа в интервале $(-5;\\;4)$: $-4,-3,-2,-1,0,1,2,3$, и ещё $x=5$. +
Наименьшее целое решение — $x=-4$. +
Ответ: наименьшее целое решение $x = -4$
` + }, + { + text: `Постройте графики уравнений системы + $$\\begin{cases} 3x + y = 5, \\\\[4pt] y - x^2 = 1 \\end{cases}$$ + и найдите сумму ординат точек пересечения.`, + sol: `Метод решения: точки пересечения графиков — это решения системы. Используем метод подстановки. +
Шаг 1. Из первого уравнения выразим $y$: $y = 5 - 3x$. +
Шаг 2. Подставим во второе уравнение $y - x^2 = 1$: +$$(5-3x)-x^2 = 1 \\;\\implies\\; -x^2 - 3x + 4 = 0 \\;\\implies\\; x^2+3x-4=0$$ +Шаг 3. По теореме Виета ищем корни: сумма $-3$, произведение $-4$ — это $-4$ и $1$: +$$(x+4)(x-1)=0 \\;\\implies\\; x_1 = -4,\\; x_2 = 1$$ +Шаг 4. Находим $y$ по формуле $y = 5 - 3x$: +
при $x = -4$: $y = 5 - 3\\cdot(-4) = 5 + 12 = 17$; +
при $x = 1$: $y = 5 - 3\\cdot 1 = 2$. + + + + + + + + −4 + −3 + −2 + −1 + 1 + 17 + 2 + 1 + + + + + (−4; 17) + (1; 2) + y=5−3x + y=x²+1 + x + y + +Шаг 5. Точки пересечения: $(-4;\\,17)$ и $(1;\\,2)$. Сумма ординат: +$$17 + 2 = 19$$ +
Ответ: $19$
` + }, + { + text: `Три числа, дающие в сумме $18$, являются последовательными членами арифметической прогрессии. + Если от первого числа вычесть $2$, от второго вычесть $3$, а третье число оставить без изменения, + то полученные числа будут последовательными членами геометрической прогрессии. + Найдите эти числа.`, + sol: `Пусть три члена АП: $6-d$, $6$, $6+d$ (сумма $=3\\cdot6=18$). +
После преобразований получаем три члена ГП: $(4-d)$, $3$, $(6+d)$. +
Условие ГП ($b^2=ac$): +$$3^2=(4-d)(6+d) \\implies 9=24+4d-6d-d^2$$ +$$9=24-2d-d^2 \\implies d^2+2d-15=0$$ +$$D=4+60=64=8^2 \\implies d=\\dfrac{-2\\pm8}{2}$$ +$d=3$: АП: $3,\\ 6,\\ 9$. ГП: $1,\\ 3,\\ 9$ (знаменатель $3$) ✓ +
$d=-5$: АП: $11,\\ 6,\\ 1$. ГП: $9,\\ 3,\\ 1$ (знаменатель $\\tfrac{1}{3}$) ✓ +
Ответ: $3,\\ 6,\\ 9$ или $1,\\ 6,\\ 11$
` + }, + { + text: `$CK$ — медиана треугольника $ABC$, площадь которого $240$ см². + Точка $E$ — середина медианы $CK$. + Луч $AE$ пересекает сторону $BC$ в точке $M$. + Найдите площадь четырёхугольника $КЕМВ$.`, + sol: ` + + + + + + + + + C + A + B + K + E + M + КЕМВ + +Шаг 1. Медиана $CK$ делит $\\triangle ABC$ пополам ($K$ — середина $AB$): +$$S_{ACK} = S_{BCK} = 120\\text{ см}^2$$ +Шаг 2. Находим точку $M$ на $BC$. +
Координаты: $A=(0,0)$, $B=(2,0)$, $C=(0,h)$, $K=(1,0)$, $E=\bigl(\tfrac{1}{2},\tfrac{h}{2}\bigr)$. +
Прямая $AE$: $y=hx$. Прямая $BC$: $hx+2y=2h$. +
Подставляем $y=hx$: $3hx=2h \\implies x=\\dfrac{2}{3}$, $y=\\dfrac{2h}{3}$. +Шаг 3. $S_{\\triangle KEM}$ (вершины $K=(1,0)$, $E=(\tfrac{1}{2},\tfrac{h}{2})$, $M=(\tfrac{2}{3},\tfrac{2h}{3})$): +$$S_{KEM}=\\dfrac{1}{2}\\left|\\left(-\\dfrac{1}{2}\\right)\\cdot\\dfrac{2h}{3}-\\dfrac{h}{2}\\cdot\\left(-\\dfrac{1}{3}\\right)\\right|=\\dfrac{1}{2}\\cdot\\dfrac{h}{6}=\\dfrac{h}{12}$$ +Шаг 4. $S_{\\triangle BKM}$: основание $BK=1$, высота из $M$ на $BK$ равна $\\dfrac{2h}{3}$: +$$S_{BKM}=\\dfrac{1}{2}\cdot1\\cdot\\dfrac{2h}{3}=\\dfrac{h}{3}$$ +Шаг 5. $S_{\\triangle ABC}=h=240$. +$$S_{\\text{КЕМВ}}=S_{KEM}+S_{BKM}=\\dfrac{240}{12}+\\dfrac{240}{3}=20+80=100\\text{ см}^2$$ +
Ответ: $100$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v23.js b/frontend/js/exam9/variants/v23.js new file mode 100644 index 0000000..58ef3d5 --- /dev/null +++ b/frontend/js/exam9/variants/v23.js @@ -0,0 +1,202 @@ +VARIANTS[23] = { + label: "Вариант 23", + tasks: [ + { + text: `Какая из следующих точек является центром окружности, + заданной уравнением $(x-3)^2 + (y+1)^2 = 1$:`, + opts: [ + ["а", "$A(-3;\\;1)$"], ["б", "$B(-3;\\;{-1})$"], ["в", "$C(3;\\;{-1})$"], + ["г", "$D(3;\\;1)$"], ["д", "$E(-1;\\;3)$"], + ], + sol: `Стандартное уравнение окружности: $(x-a)^2+(y-b)^2=R^2$, центр $= (a,\\,b)$. +
Сравниваем: $(x-3)^2+(y+1)^2=1$ — здесь $a=3$, $b=-1$. +
Ответ: в) $C(3;\\;{-1})$
` + }, + { + text: `Произведение дробей $\\dfrac{a}{a+b} \\cdot \\dfrac{a+b}{b}$ равно:`, + opts: [ + ["а", "$\\dfrac{a}{b}$"], ["б", "$\\dfrac{a+b}{b}$"], ["в", "$\\dfrac{a+b}{a}$"], + ["г", "$1$"], ["д", "$a+b$"], + ], + sol: `Сокращаем множитель $(a+b)$: +$$\\dfrac{a}{a+b}\\cdot\\dfrac{a+b}{b} = \\dfrac{a\\cdot(a+b)}{(a+b)\\cdot b} = \\dfrac{a}{b}$$ +
Ответ: а) $\\dfrac{a}{b}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "через точку, лежащую на окружности, можно провести только одну касательную к этой окружности;"], + ["б", "площадь прямоугольного треугольника равна произведению катетов;"], + ["в", "если два угла одного треугольника равны $20^{\\circ}$ и $80^{\\circ}$, другого — $80^{\\circ}$ и $20^{\\circ}$, то треугольники подобны между собой;"], + ["г", "$\\cos 30^{\\circ} = \\dfrac{\\sqrt{3}}{2}$?"], + ], + sol: `
    +
  • а) Единственная касательная через точку на окружности — верно
    • +
  • б) «Площадь прямоугольного треугольника равна произведению катетов» — НЕВЕРНО. Правильная формула: $S = \\dfrac{1}{2}\\cdot a\\cdot b$ (половина произведения катетов). Без множителя $\\tfrac{1}{2}$ формула неверна.
    • +
  • в) Одинаковые наборы углов ($20°, 80°, 80°$) ⟹ треугольники подобны — верно
    • +
  • г) $\\cos30° = \\dfrac{\\sqrt{3}}{2}$ — верно
    • +
+
Ответ: б)
` + }, + { + text: `Определите, принадлежит ли точка $A(1;\\;{-1})$ графику функции $y = 4x + 3$. + Ответ обоснуйте.`, + sol: `Подставляем координаты точки $A(1;\\,-1)$ в уравнение $y=4x+3$: +$$y = 4\\cdot1+3 = 7$$ +Получили $y=7$, но у точки $A$ координата $y=-1$. +
Так как $-1\\neq 7$, точка $A(1;\\,-1)$ не принадлежит графику. +
Ответ: нет, не принадлежит
` + }, + { + text: `Учащемуся в возрасте $12$ лет требуется не менее $8$–$9$ часов сна. + Выполняется ли это требование, если он спит $\\dfrac{1}{3}$ суток? Ответ обоснуйте.`, + sol: `Правило нахождения части от числа: чтобы найти $\\dfrac{m}{n}$ от числа $A$, нужно умножить $A$ на $\\dfrac{m}{n}$. Длительность суток: $24$ часа. +
Шаг 1. Найдём, сколько часов составляет $\\dfrac{1}{3}$ суток: +$$\\dfrac{1}{3}\\cdot 24 = \\dfrac{24}{3} = 8\\text{ ч}.$$ +Шаг 2. Сравним полученное время с нижней границей нормы. По условию требуется не менее $8$ часов сна, значит нужно, чтобы фактическое время сна было $\\geq 8$ ч. +
Так как $8\\geq 8$, требование выполняется: продолжительность сна равна нижней границе нормы. +
Ответ: да, требование выполняется — ровно $8$ часов сна
` + }, + { + text: `Сравните значение выражения + $\\dfrac{16}{7} \\cdot \\left(-\\dfrac{7}{8}\\right) - \\dfrac{1}{4} \\cdot \\left(-\\dfrac{5}{8}\\right) + \\dfrac{1}{5} \\cdot \\left(-\\dfrac{2}{5}\\right)$ + с числом $-2{,}5$.`, + sol: `Порядок действий: сначала выполняем умножения, затем сложение и вычитание. Правило умножения дробей: $\\dfrac{a}{b}\\cdot\\dfrac{c}{d}=\\dfrac{ac}{bd}$. Знак произведения: минус на плюс даёт минус, минус на минус — плюс. +
Шаг 1. Вычислим каждое произведение по отдельности. +$$\\dfrac{16}{7}\\cdot\\left(-\\dfrac{7}{8}\\right) = -\\dfrac{16\\cdot 7}{7\\cdot 8} = -\\dfrac{16}{8} = -2;$$ +$$\\dfrac{1}{4}\\cdot\\left(-\\dfrac{5}{8}\\right) = -\\dfrac{5}{32}\\;\\;\\text{(минус на плюс — минус)};$$ +$$\\dfrac{1}{5}\\cdot\\left(-\\dfrac{2}{5}\\right) = -\\dfrac{2}{25}.$$ +Шаг 2. Подставим найденные произведения. По условию второе слагаемое идёт со знаком «минус», поэтому $-\\left(-\\dfrac{5}{32}\\right)=+\\dfrac{5}{32}$: +$$-2 + \\dfrac{5}{32} - \\dfrac{2}{25}.$$ +Шаг 3. Приведём дроби $\\dfrac{5}{32}$ и $\\dfrac{2}{25}$ к общему знаменателю $800$ (это наименьшее общее кратное $32$ и $25$): +$$\\dfrac{5}{32}=\\dfrac{125}{800},\\qquad \\dfrac{2}{25}=\\dfrac{64}{800}.$$ +$$-2 + \\dfrac{125}{800} - \\dfrac{64}{800} = -2 + \\dfrac{61}{800}.$$ +Шаг 4. Запишем $-2$ как $-\\dfrac{1600}{800}$ и сложим: +$$-\\dfrac{1600}{800} + \\dfrac{61}{800} = -\\dfrac{1539}{800}.$$ +Шаг 5. Сравним $-\\dfrac{1539}{800}$ и $-2{,}5=-\\dfrac{2000}{800}$. Из двух отрицательных чисел больше то, у которого модуль меньше: +$$|-1539|=1539\\lt 2000=|-2000| \\;\\implies\\; -\\dfrac{1539}{800}\\gt -\\dfrac{2000}{800}.$$ +Значит, значение выражения больше, чем $-2{,}5$. +
Ответ: выражение $\\gt -2{,}5$
` + }, + { + text: `Найдите площадь треугольника со сторонами $9$ см, $12$ см и $15$ см.`, + sol: `Теорема, обратная теореме Пифагора: если для сторон $a$, $b$, $c$ треугольника выполняется $a^2+b^2=c^2$, то треугольник прямоугольный, $c$ — гипотенуза. +
Формула площади прямоугольного треугольника: $S = \\dfrac{1}{2}\\cdot a\\cdot b$, где $a$ и $b$ — катеты. +
Шаг 1. Проверим, является ли треугольник прямоугольным. Для этого квадрат большей стороны сравниваем с суммой квадратов двух других: +$$9^2+12^2 = 81+144 = 225$$ +$$15^2 = 225$$ +Поскольку $9^2+12^2=15^2$, треугольник прямоугольный, катеты — $9$ и $12$, гипотенуза — $15$. + + + + A + B + C + 12 см + 9 см + 15 см + +Шаг 2. Применяем формулу площади: +$$S = \\dfrac{1}{2}\\cdot 9\\cdot 12 = \\dfrac{108}{2} = 54\\text{ см}^2$$ +
Ответ: $54$ см²
` + }, + { + text: `Найдите частное $a$ и $b$, если + $a = 3^6 \\cdot (5^{-1})^{-2} \\cdot \\dfrac{1}{4^{-2}}$ + и $b = 3^8 \\cdot 5^3 \\cdot \\dfrac{1}{4^{-1}}$.`, + sol: `Свойства степеней: $(x^m)^n=x^{mn}$,  $\\dfrac{1}{x^{-n}}=x^n$,  $\\dfrac{x^m}{x^n}=x^{m-n}$,  $x^{-n}=\\dfrac{1}{x^n}$. +
Шаг 1. Упростим выражение $a$. По правилу степени степени $(5^{-1})^{-2}=5^{(-1)\\cdot(-2)}=5^2$, а $\\dfrac{1}{4^{-2}}=4^2$: +$$a = 3^6\\cdot 5^2\\cdot 4^2.$$ +Шаг 2. Упростим $b$. Аналогично $\\dfrac{1}{4^{-1}}=4^1=4$: +$$b = 3^8\\cdot 5^3\\cdot 4.$$ +Шаг 3. Найдём частное. Сгруппируем одинаковые основания и применим правило $\\dfrac{x^m}{x^n}=x^{m-n}$: +$$\\dfrac{a}{b} = \\dfrac{3^6\\cdot 5^2\\cdot 4^2}{3^8\\cdot 5^3\\cdot 4} = 3^{6-8}\\cdot 5^{2-3}\\cdot 4^{2-1} = 3^{-2}\\cdot 5^{-1}\\cdot 4.$$ +Шаг 4. Запишем отрицательные степени как дроби: +$$3^{-2}\\cdot 5^{-1}\\cdot 4 = \\dfrac{4}{3^2\\cdot 5} = \\dfrac{4}{9\\cdot 5} = \\dfrac{4}{45}.$$ +
Ответ: $\\dfrac{a}{b}=\\dfrac{4}{45}$
` + }, + { + text: `Бобруйскому заводу тракторных деталей и агрегатов поступил заказ + на изготовление $800$ малогабаритных прицепов для трактора к определённому сроку. + Работая точно по графику, рабочие выполнили $25\\%$ заказа, + а затем стали собирать на $10$ прицепов больше и выполнили заказ за $2$ дня + до назначенного срока. За сколько дней рабочие выполнили заказ?`, + sol: `Пусть плановая выработка $= x$ прицепов в день. +
По плану: весь заказ занял бы $\\dfrac{800}{x}$ дней. +
Фактически: +
    +
  • $25\\%$ от $800 = 200$ прицепов — сделали за $\\dfrac{200}{x}$ дней по плану
    • +
  • Оставшиеся $600$ — делали с темпом $(x+10)$ за $\\dfrac{600}{x+10}$ дней
    • +
+Закончили на $2$ дня раньше: +$$\\dfrac{800}{x} - \\left(\\dfrac{200}{x}+\\dfrac{600}{x+10}\\right) = 2$$ +$$\\dfrac{600}{x} - \\dfrac{600}{x+10} = 2$$ +Умножаем на $x(x+10)$: +$$600(x+10)-600x = 2x(x+10)$$ +$$6000 = 2x^2+20x$$ +$$x^2+10x-3000=0$$ +$$D = 100+12000 = 12100 = 110^2$$ +$$x = \\dfrac{-10+110}{2} = 50$$ +Плановый срок: $\\dfrac{800}{50}=16$ дней. Фактически: $16-2=14$ дней. +
Проверка: $\\dfrac{200}{50}+\\dfrac{600}{60} = 4+10 = 14$ дней ✓ +
Ответ: $14$ дней
` + }, + { + text: `В окружности проведены две взаимно перпендикулярные хорды $AB$ и $CD$, + которые пересекаются в точке $K$, $AK = 2$, $KB = 6$, $DK = 3$. + Найдите площадь круга, ограниченного этой окружностью.`, + sol: `Шаг 1. Находим $CK$ — теорема о пересекающихся хордах. +
При пересечении хорд произведения отрезков равны: $AK\\cdot KB = CK\\cdot KD$: +$$2\\cdot6 = CK\\cdot3 \\implies CK = 4$$ +Длины хорд: $AB = 2+6 = 8$ см, $CD = 4+3 = 7$ см. +
Шаг 2. Серединные перпендикуляры проходят через центр. +
Обозначим $M$ — середина $AB$, $N$ — середина $CD$. По свойству хорд: +$$OM \\perp AB \\quad \\text{и} \\quad ON \\perp CD$$ +Шаг 3. Строим прямоугольник $ONKM$. +
Так как $AB \\perp CD$: +
    +
  • $OM\\perp AB$ и $AB\\perp CD$ ⟹ $OM\\parallel CD$
    • +
  • $ON\\perp CD$ и $AB\\perp CD$ ⟹ $ON\\parallel AB$
    • +
+Четырёхугольник $ONKM$ — прямоугольник. Его стороны: +$$KM = AM - AK = 4 - 2 = 2$$ +Точка $N$ — середина $CD$, $CN=\\dfrac{7}{2}=3{,}5$. Так как $CK=4>3{,}5$, то $N$ лежит между $C$ и $K$: +$$KN = CK - CN = 4 - 3{,}5 = 0{,}5 = \\dfrac{1}{2}$$ +Противоположные стороны прямоугольника равны: +$$ON = KM = 2, \\qquad OM = KN = \\dfrac{1}{2}$$ + + + + + + + + + + + + A + B + C + D + K + O + M + N + 2 + 6 + 4 + 3 + KM=2 + OM=½ + R + +Шаг 4. Находим радиус по теореме Пифагора. +
В прямоугольном треугольнике $OMA$ (прямой угол у $M$, $OA = R$): +$$R^2 = OM^2 + AM^2 = \\left(\\dfrac{1}{2}\\right)^2 + 4^2 = \\dfrac{1}{4} + 16 = \\dfrac{65}{4}$$ +Шаг 5. Площадь круга. +$$S = \\pi R^2 = \\dfrac{65\\pi}{4}\\text{ см}^2$$ +
Ответ: $\\dfrac{65\\pi}{4}$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v24.js b/frontend/js/exam9/variants/v24.js new file mode 100644 index 0000000..7214d46 --- /dev/null +++ b/frontend/js/exam9/variants/v24.js @@ -0,0 +1,191 @@ +VARIANTS[24] = { + label: "Вариант 24", + tasks: [ + { + text: `Какая из следующих точек является центром окружности, + заданной уравнением $(x+4)^2 + (y-3)^2 = 2$:`, + opts: [ + ["а", "$A(-4;\\;3)$"], ["б", "$B(-4;\\;{-3})$"], ["в", "$C(4;\\;{-3})$"], + ["г", "$D(4;\\;3)$"], ["д", "$E(3;\\;{-4})$"], + ], + sol: `Стандартное уравнение окружности: $(x-a)^2+(y-b)^2=R^2$, центр $= (a,\\,b)$. +
Сравниваем: $(x+4)^2+(y-3)^2=2$ — здесь $a=-4$, $b=3$. +
Ответ: а) $A(-4;\\;3)$
` + }, + { + text: `Произведение дробей $\\dfrac{1}{a-b} \\cdot \\dfrac{a^2-b^2}{1}$ равно:`, + opts: [ + ["а", "$a+b$"], ["б", "$\\dfrac{a+b}{b}$"], ["в", "$\\dfrac{a+b}{a}$"], + ["г", "$1$"], ["д", "$\\dfrac{1}{a+b}$"], + ], + sol: `Разложим $a^2-b^2=(a-b)(a+b)$ и сократим $(a-b)$: +$$\\dfrac{1}{a-b}\\cdot\\dfrac{a^2-b^2}{1} = \\dfrac{(a-b)(a+b)}{a-b} = a+b$$ +
Ответ: а) $a+b$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "через точку, лежащую вне окружности, можно провести ровно две касательные к этой окружности;"], + ["б", "площадь ромба равна произведению диагоналей;"], + ["в", "если стороны одного треугольника равны $6$, $8$ и $10$, другого — $3$, $4$ и $5$, то треугольники подобны между собой;"], + ["г", "$\\cos 60^{\\circ} = \\dfrac{1}{2}$?"], + ], + sol: `
    +
  • а) Через внешнюю точку окружности проходят ровно две касательные — верно
    • +
  • б) «Площадь ромба равна произведению диагоналей» — НЕВЕРНО. Правильная формула: $S=\\dfrac{1}{2}d_1 d_2$ (половина произведения диагоналей).
    • +
  • в) Стороны $6:8:10=3:4:5$, коэффициент подобия $2$ — верно
    • +
  • г) $\\cos60°=\\dfrac{1}{2}$ — верно
    • +
+
Ответ: б)
` + }, + { + text: `Определите, принадлежит ли точка $A(-1;\\;1)$ графику функции $y = 3x + 4$. + Ответ обоснуйте.`, + sol: `Подставляем координаты точки $A(-1;\\,1)$ в уравнение $y=3x+4$: +$$y = 3\\cdot(-1)+4 = -3+4 = 1$$ +Получили $y=1$, и у точки $A$ координата $y=1$. +
Так как $1=1$, точка $A(-1;\\,1)$ принадлежит графику функции. +
Ответ: да, принадлежит
` + }, + { + text: `Учащемуся в возрасте $10$ лет требуется не менее $9$–$10$ часов сна. + Выполняется ли это требование, если он спит $\\dfrac{5}{12}$ суток? Ответ обоснуйте.`, + sol: `Правило нахождения части от числа: чтобы найти $\\dfrac{m}{n}$ от числа $A$, нужно умножить $A$ на $\\dfrac{m}{n}$. Длительность суток: $24$ часа. +
Шаг 1. Найдём, сколько часов составляет $\\dfrac{5}{12}$ суток: +$$\\dfrac{5}{12}\\cdot 24 = \\dfrac{5\\cdot 24}{12} = \\dfrac{120}{12} = 10\\text{ ч}.$$ +Шаг 2. Сравним полученное время с нижней границей нормы. По условию требуется не менее $9$ часов, значит нужно, чтобы фактическое время сна было $\\geq 9$ ч. +
Так как $10\\geq 9$, требование выполняется. +
Ответ: да, требование выполняется — ровно $10$ часов сна
` + }, + { + text: `Сравните значение выражения + $\\dfrac{5}{12} \\cdot \\left(-\\dfrac{3}{5}\\right) + \\dfrac{1}{4} \\cdot \\left(-\\dfrac{3}{8}\\right) - \\dfrac{1}{7} \\cdot \\dfrac{4}{7}$ + с числом $-2$.`, + sol: `Порядок действий: сначала умножения, потом сложение и вычитание. Правило умножения дробей: $\\dfrac{a}{b}\\cdot\\dfrac{c}{d}=\\dfrac{ac}{bd}$. Знак произведения: минус на плюс — минус, плюс на плюс — плюс. +
Шаг 1. Вычислим каждое произведение, сокращая по ходу: +$$\\dfrac{5}{12}\\cdot\\left(-\\dfrac{3}{5}\\right) = -\\dfrac{5\\cdot 3}{12\\cdot 5} = -\\dfrac{3}{12} = -\\dfrac{1}{4};$$ +$$\\dfrac{1}{4}\\cdot\\left(-\\dfrac{3}{8}\\right) = -\\dfrac{3}{32};$$ +$$\\dfrac{1}{7}\\cdot\\dfrac{4}{7} = \\dfrac{4}{49}.$$ +Шаг 2. Подставим найденные значения в исходное выражение: +$$-\\dfrac{1}{4} + \\left(-\\dfrac{3}{32}\\right) - \\dfrac{4}{49} = -\\dfrac{1}{4} - \\dfrac{3}{32} - \\dfrac{4}{49}.$$ +Шаг 3. Найдём общий знаменатель. НОК$(4;32;49) = 32\\cdot 49 = 1568$. Приведём дроби: +$$-\\dfrac{1}{4} = -\\dfrac{392}{1568},\\quad -\\dfrac{3}{32} = -\\dfrac{147}{1568},\\quad -\\dfrac{4}{49} = -\\dfrac{128}{1568}.$$ +Шаг 4. Сложим числители: +$$-\\dfrac{392+147+128}{1568} = -\\dfrac{667}{1568}\\approx -0{,}43.$$ +Шаг 5. Сравним с числом $-2$. Так как $-0{,}43\\gt -2$ (отрицательное число с меньшим модулем больше), получаем: +
Ответ: выражение $\\gt -2$
` + }, + { + text: `Найдите площадь треугольника со сторонами $10$ см, $24$ см и $26$ см.`, + sol: `Теорема, обратная теореме Пифагора: если для сторон $a$, $b$, $c$ треугольника выполнено равенство $a^2+b^2=c^2$, то треугольник прямоугольный, а $c$ — гипотенуза. +
Формула площади прямоугольного треугольника: $S = \\dfrac{1}{2}\\cdot a\\cdot b$, где $a$ и $b$ — катеты. +
Шаг 1. Проверим, прямоугольный ли треугольник. Сравним квадрат большей стороны с суммой квадратов двух других: +$$10^2+24^2 = 100+576 = 676$$ +$$26^2 = 676$$ +Поскольку $10^2+24^2=26^2$, треугольник прямоугольный, его катеты — $10$ и $24$. + + + + A + B + C + 24 см + 10 см + 26 см + +Шаг 2. Применяем формулу площади: +$$S = \\dfrac{1}{2}\\cdot 10\\cdot 24 = 120\\text{ см}^2$$ +
Ответ: $120$ см²
` + }, + { + text: `Найдите частное $m$ и $n$, если + $m = 3^{-2} \\cdot 3^2 \\cdot \\dfrac{(7^{-1})^{-2}}{3^{-5}}$ + и $n = \\dfrac{1}{7^{-2}} \\cdot 9^{-1} \\cdot \\dfrac{1}{3^{-5}}$.`, + sol: `Свойства степеней: $x^m\\cdot x^n=x^{m+n}$,  $(x^m)^n=x^{mn}$,  $\\dfrac{x^m}{x^n}=x^{m-n}$,  $\\dfrac{1}{x^{-n}}=x^n$. +
Шаг 1. Упростим $m$. По свойству степени степени $(7^{-1})^{-2}=7^2$, а деление на $3^{-5}$ равно умножению на $3^5$: +$$m = 3^{-2}\\cdot 3^2\\cdot 7^2\\cdot 3^5 = 3^{-2+2+5}\\cdot 7^2 = 3^5\\cdot 7^2.$$ +Шаг 2. Упростим $n$. Имеем $\\dfrac{1}{7^{-2}}=7^2$, $9^{-1}=3^{-2}$ (так как $9=3^2$), $\\dfrac{1}{3^{-5}}=3^5$: +$$n = 7^2\\cdot 3^{-2}\\cdot 3^5 = 7^2\\cdot 3^{-2+5} = 7^2\\cdot 3^3.$$ +Шаг 3. Найдём частное. Сокращаем $7^2$ и применяем правило $\\dfrac{x^m}{x^n}=x^{m-n}$: +$$\\dfrac{m}{n} = \\dfrac{3^5\\cdot 7^2}{3^3\\cdot 7^2} = 3^{5-3} = 3^2 = 9.$$ +
Ответ: $\\dfrac{m}{n}=9$
` + }, + { + text: `Барановичскому станкостроительному заводу поступил заказ на изготовление $1200$ дробилок, + которые используют для дробления пластиковых деталей к определённому сроку. + Работая точно по графику, рабочие изготовили $25\\%$ заказа, + а затем стали производить в день на $50$ дробилок больше и выполнили заказ за $3$ дня + до назначенного срока. За сколько дней рабочие выполнили заказ?`, + sol: `Метод введения переменной: обозначим неизвестную плановую выработку буквой $x$ и составим уравнение по условию о сроках. +
Шаг 1. Пусть $x$ — количество дробилок, которое нужно было выпускать в день по плану. Тогда плановый срок выполнения заказа: $\\dfrac{1200}{x}$ дней. +
Шаг 2. Сначала рабочие изготовили $25\\%$ заказа, то есть $\\dfrac{25}{100}\\cdot 1200 = 300$ дробилок. По плановой выработке это заняло $\\dfrac{300}{x}$ дней. +
Шаг 3. Оставшиеся $1200-300=900$ дробилок делали по $x+50$ штук в день. Это заняло $\\dfrac{900}{x+50}$ дней. +
Шаг 4. Закончили на $3$ дня раньше плана, значит фактический срок на $3$ меньше планового: +$$\\dfrac{1200}{x} - \\left(\\dfrac{300}{x}+\\dfrac{900}{x+50}\\right) = 3.$$ +Упрощаем левую часть, $\\dfrac{1200}{x}-\\dfrac{300}{x}=\\dfrac{900}{x}$: +$$\\dfrac{900}{x} - \\dfrac{900}{x+50} = 3.$$ +Шаг 5. Умножим обе части на $x(x+50)$, чтобы избавиться от знаменателей: +$$900(x+50) - 900x = 3x(x+50);$$ +$$45000 = 3x^2+150x \\;\\implies\\; x^2+50x-15000 = 0.$$ +Шаг 6. Решим квадратное уравнение через дискриминант: +$$D = 50^2 + 4\\cdot 15000 = 2500+60000 = 62500 = 250^2;$$ +$$x = \\dfrac{-50+250}{2} = 100\\;\\;\\text{(второй корень отрицательный, не подходит)}.$$ +Шаг 7. Найдём фактический срок. По плану заказ занял бы $\\dfrac{1200}{100}=12$ дней, фактически — на $3$ дня меньше: $12-3=9$ дней. +
Проверка: первые $\\dfrac{300}{100}=3$ дня + последние $\\dfrac{900}{150}=6$ дней = $9$ дней. +
Ответ: $9$ дней
` + }, + { + text: `В окружности проведены две взаимно перпендикулярные хорды $AB$ и $CD$, + которые пересекаются в точке $K$, $AK = 4$, $KB = 9$, $DK = 3$. + Найдите площадь круга, ограниченного этой окружностью.`, + sol: `Теорема о пересекающихся хордах: если две хорды пересекаются в точке $K$, то $AK\\cdot KB = CK\\cdot KD$. +
Свойство хорды: серединный перпендикуляр к хорде проходит через центр окружности. +
Теорема Пифагора: $c^2 = a^2+b^2$ для прямоугольного треугольника. +
Формула площади круга: $S=\\pi R^2$. +
Шаг 1. По теореме о пересекающихся хордах: +$$AK\\cdot KB = CK\\cdot KD \\;\\implies\\; 4\\cdot 9 = CK\\cdot 3 \\;\\implies\\; CK = 12$$ +Длины хорд: $AB = AK+KB = 4+9 = 13$, $CD = CK+KD = 12+3 = 15$. +
Шаг 2. Пусть $M$ — середина $AB$, $N$ — середина $CD$. Тогда $OM\\perp AB$ и $ON\\perp CD$ (по свойству перпендикуляра из центра к хорде). +
Шаг 3. Так как $AB\\perp CD$ и $OM\\perp AB$, $ON\\perp CD$, четырёхугольник $ONKM$ — прямоугольник, поэтому $OM=KN$ и $ON=KM$. +
Шаг 4. Находим $AM$ и $KM$: +$$AM = \\dfrac{AB}{2} = \\dfrac{13}{2} = 6{,}5$$ +$$KM = AM - AK = 6{,}5 - 4 = 2{,}5$$ +Шаг 5. Находим $CN$ и $KN$: +$$CN = \\dfrac{CD}{2} = \\dfrac{15}{2} = 7{,}5$$ +$$KN = CK - CN = 12 - 7{,}5 = 4{,}5$$ +Значит $OM=KN=4{,}5$. + + + + + + + + + + + + A + B + C + D + K + O + M + N + 4 + 9 + 12 + 3 + KM=2,5 + OM=4,5 + R + +Шаг 6. В прямоугольном треугольнике $OMA$ (прямой угол при $M$): $OA = R$, $OM = 4{,}5$, $AM = 6{,}5$. По теореме Пифагора: +$$R^2 = OM^2 + AM^2 = 4{,}5^2 + 6{,}5^2 = 20{,}25 + 42{,}25 = 62{,}5 = \\dfrac{125}{2}$$ +Шаг 7. Площадь круга: +$$S = \\pi R^2 = \\dfrac{125\\pi}{2}\\text{ см}^2$$ +
Ответ: $\\dfrac{125\\pi}{2}$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v25.js b/frontend/js/exam9/variants/v25.js new file mode 100644 index 0000000..9ca7e27 --- /dev/null +++ b/frontend/js/exam9/variants/v25.js @@ -0,0 +1,226 @@ +VARIANTS[25] = { + label: "Вариант 25", + tasks: [ + { + text: `Какое из приведённых ниже выражений тождественно равно произведению $-2(-5-a)$:`, + opts: [ + ["а", "$10-a$"], ["б", "$2a-10$"], ["в", "$10+2a$"], + ["г", "$10-2a$"], ["д", "$2a-5$"], + ], + sol: `Раскрываем скобки: +$$-2(-5-a) = (-2)\\cdot(-5) + (-2)\\cdot(-a) = 10 + 2a$$ +
Ответ: в) $10+2a$
` + }, + { + text: `Значение выражения $68{,}3 - 50{,}08$ равно:`, + opts: [ + ["а", "$8{,}38$"], ["б", "$8{,}22$"], ["в", "$18{,}32$"], + ["г", "$7{,}32$"], ["д", "$18{,}22$"], + ], + sol: `$$68{,}30 - 50{,}08 = 18{,}22$$ +
Ответ: д) $18{,}22$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "треугольник со сторонами $3$, $4$, $5$ — прямоугольный;"], + ["б", "центр описанной окружности треугольника лежит на пересечении серединных перпендикуляров к его сторонам;"], + ["в", "если у параллелограмма диагонали равны, то это прямоугольник;"], + ["г", "вписанный угол равен соответствующему центральному углу?"], + ], + sol: `
    +
  • а) $3^2+4^2=25=5^2$ ⟹ прямоугольный — верно
    • +
  • б) Центр описанной окружности — точка пересечения серединных перпендикуляров — верно
    • +
  • в) Если диагонали параллелограмма равны — это прямоугольник — верно
    • +
  • г) «Вписанный угол равен центральному» — НЕВЕРНО. Вписанный угол равен половине центрального угла, опирающегося на ту же дугу.
    • +
+
Ответ: г)
` + }, + { + text: `Последовательность $-18;\\; -16;\\; -14;\\; \\ldots$ — арифметическая прогрессия. + Продолжите её далее, записав ещё три члена прогрессии.`, + sol: `Разность прогрессии: $d = -16-(-18) = 2$. +
Продолжаем, прибавляя $2$ к каждому члену: +$$-14+2=-12; \\ -12+2=-10; \\ -10+2=-8$$ +
Ответ: $-12;\;-10;\;-8$
` + }, + { + text: `В треугольнике две стороны равны $6$ см и $10$ см, + а сумма углов, противолежащих этим сторонам, равна $120^{\\circ}$. + Найдите третью сторону треугольника.`, + sol: `Пусть $a=6$, $b=10$ — данные стороны, $\\alpha$ и $\\beta$ — углы, им противолежащие. +
По условию: $\\alpha+\\beta=120°$. Значит, третий угол: +$$\\gamma = 180°-120°=60°$$ +Угол $\\gamma$ заключён между сторонами $a=6$ и $b=10$. + + + + C + A + B + 60° + 6 + 10 + $c$ + +По теореме косинусов (для стороны $c$, противолежащей углу $\\gamma=60°$): +$$c^2 = a^2+b^2-2ab\\cos\\gamma = 6^2+10^2-2\\cdot6\\cdot10\\cdot\\cos60°$$ +$$c^2 = 36+100-120\\cdot\\tfrac{1}{2} = 136-60 = 76$$ +$$c = \\sqrt{76} = 2\\sqrt{19}\\text{ см}$$ +
Ответ: $2\\sqrt{19}$ см
` + }, + { + text: `После проведения профилактических мероприятий необходимо наполнить один из бассейнов + спорткомплекса объёмом $1500$ л. + Через первый кран в бассейн вливается $30$ л воды в минуту, а через второй — $20$ л в минуту. + За какое время бассейн будет наполнен, если открыть оба крана одновременно?`, + sol: `Правило совместной работы: при одновременной работе производительности (скорости наполнения) складываются. Формула времени: $t=\\dfrac{V}{v}$, где $V$ — объём, $v$ — суммарная производительность. +
Шаг 1. Найдём суммарную производительность двух кранов. Так как краны работают одновременно, объёмы воды, поступающие в минуту, складываются: +$$v = 30 + 20 = 50\\text{ л/мин}.$$ +Шаг 2. Делим объём бассейна на совместную производительность, чтобы найти время наполнения: +$$t = \\dfrac{V}{v} = \\dfrac{1500}{50} = 30\\text{ мин}.$$ +
Ответ: $30$ минут
` + }, + { + text: `Найдите наибольшее целое число, принадлежащее множеству решений системы неравенств + $$\\begin{cases} \\dfrac{1}{3}(x+3) \\geq \\dfrac{6x-7}{4}, \\\\[6pt] \\dfrac{1}{4}x + 3 \\leq 6{,}5x + 2. \\end{cases}$$`, + sol: `Метод решения системы неравенств: решаем каждое неравенство отдельно, затем берём пересечение решений. +
Шаг 1. Решаем первое неравенство. Умножим обе части на $12$ (общий знаменатель), чтобы избавиться от дробей: +$$\\dfrac{1}{3}(x+3) \\geq \\dfrac{6x-7}{4} \\;\\;\\bigg|\\cdot 12$$ +$$4(x+3) \\geq 3(6x-7)$$ +$$4x+12 \\geq 18x-21$$ +$$12+21 \\geq 18x-4x$$ +$$33 \\geq 14x \\;\\implies\\; x \\leq \\dfrac{33}{14}\\approx 2{,}36$$ +Шаг 2. Решаем второе неравенство. +$$\\dfrac{1}{4}x + 3 \\leq 6{,}5x + 2$$ +$$3-2 \\leq 6{,}5x - 0{,}25x$$ +$$1 \\leq 6{,}25x \\;\\implies\\; x \\geq \\dfrac{1}{6{,}25}=\\dfrac{4}{25}=0{,}16$$ +Шаг 3. Пересечение решений: $\\dfrac{4}{25}\\leq x\\leq\\dfrac{33}{14}$, то есть приблизительно $0{,}16\\leq x\\leq 2{,}36$. + + + + 0 + 1 + 2 + 3 + + + + 4/25 + 33/14 + + 2 + +Шаг 4. Среди целых чисел в промежутке $[0{,}16;\\,2{,}36]$ есть $1$ и $2$. Наибольшее из них — $x=2$. +
Ответ: $2$
` + }, + { + text: `Постройте график функции $y = \\dfrac{(2x-5)^2}{2x-5}$. + Определите, при каких значениях аргумента значение функции не больше $7$.`, + sol: `Правило сокращения дроби: если множитель встречается в числителе и в знаменателе, его можно сократить, но только при условии, что он не равен нулю. +
Шаг 1. Найдём ОДЗ. Знаменатель не должен равняться нулю: +$$2x-5\\neq 0 \\;\\implies\\; x\\neq\\dfrac{5}{2}.$$ +Шаг 2. Упростим выражение. В числителе $(2x-5)^2=(2x-5)\\cdot(2x-5)$, поэтому при $x\\neq\\dfrac{5}{2}$ один множитель $(2x-5)$ сокращается: +$$y = \\dfrac{(2x-5)^2}{2x-5} = 2x-5,\\quad x\\neq\\dfrac{5}{2}.$$ +Значит, график — это прямая $y=2x-5$ с выколотой точкой при $x=\\dfrac{5}{2}$, где $y=2\\cdot\\dfrac{5}{2}-5=0$. + + + + + + + + + + + + + 7 + + 1 + 6 + 5/2 + y=2x−5 + y=7 + x + y + +Шаг 3. Решаем неравенство $y\\leq 7$. Подставляем упрощённое выражение: +$$2x-5\\leq 7 \\;\\implies\\; 2x\\leq 12 \\;\\implies\\; x\\leq 6.$$ +Шаг 4. Учитываем ОДЗ — точка $x=\\dfrac{5}{2}$ выколота из графика, значит её исключаем из ответа. +
Ответ: $x\\leq 6$, $x\\neq\\dfrac{5}{2}$
` + }, + { + text: `На плане размеры прямоугольника $32$ мм $\\times$ $25$ мм. + В реальности площадь прямоугольника равна $200$ см². + Изобразите в заданном масштабе квадрат, если по реальным измерениям + его периметр на $230$ мм больше периметра прямоугольника.`, + sol: `Шаг 1. Масштаб. +
Площадь прямоугольника на плане: $32\\cdot25=800$ мм². В реальности: $200$ см² $=20000$ мм². +$$k^2=\\dfrac{20000}{800}=25 \\implies k=5$$ +Масштаб $1:5$ (1 мм на плане = 5 мм в реальности). +
Шаг 2. Периметр прямоугольника (реальный). +
Реальные размеры: $32\\cdot5=160$ мм и $25\\cdot5=125$ мм. +$$P_{\\text{пр}} = 2(160+125) = 570\\text{ мм}$$ +Шаг 3. Сторона квадрата. +$$P_{\\text{кв}} = 570+230 = 800\\text{ мм} \\implies a = \\dfrac{800}{4}=200\\text{ мм}$$ +Шаг 4. Сторона квадрата на плане. +$$a_{\\text{план}} = \\dfrac{200}{5} = 40\\text{ мм}$$ + + + Прямоугольник + 32 мм × 25 мм + + Квадрат + 40 мм × 40 мм + план (масштаб 1:5) + +
Ответ: квадрат со стороной $40$ мм на плане ($200$ мм в реальности)
` + }, + { + text: `Точка $M$ — середина стороны $BC$ квадрата $ABCD$, площадь которого равна $20$ см². + К отрезку $AM$ проведён перпендикуляр $DK$. + Найдите площадь четырёхугольника $DKMC$.`, + sol: `Пусть сторона квадрата $a$, тогда $a^2=20$. + + + + + + + + + A + B + C + D + M + K + DKMC + +Пусть сторона квадрата $a$, тогда $a^2=20$. +
Шаг 1. Квадрат разбивается отрезком $AM$ и перпендикуляром $DK$ на три части: +$$S_{ABCD} = S_{\\triangle ABM} + S_{\\triangle ADK} + S_{DKMC}$$ +Шаг 2. Находим $S_{\\triangle ABM}$. +
Прямой угол при $B$, катеты $AB=a$ и $BM=\\dfrac{a}{2}$: +$$S_{\\triangle ABM}=\\dfrac{1}{2}\\cdot a\\cdot\\dfrac{a}{2}=\\dfrac{a^2}{4}=5\\text{ см}^2$$ +Шаг 3. Находим $AM$. +
В прямоугольном $\\triangle ABM$ по теореме Пифагора: +$$AM=\\sqrt{AB^2+BM^2}=\\sqrt{a^2+\\dfrac{a^2}{4}}=\\dfrac{a\\sqrt{5}}{2}$$ +Шаг 4. Находим $S_{\\triangle ADM}$. +
Основание $AD=a$, высота из $M$ на $AD$ = ширина квадрата $= a$: +$$S_{\\triangle ADM}=\\dfrac{1}{2}\\cdot a\\cdot a=\\dfrac{a^2}{2}=10\\text{ см}^2$$ +Шаг 5. Находим $DK$. +
$DK$ — высота треугольника $ADM$, проведённая к основанию $AM$: +$$S_{\\triangle ADM}=\\dfrac{1}{2}\\cdot AM\\cdot DK \\implies DK=\\dfrac{2\\cdot S_{\\triangle ADM}}{AM}=\\dfrac{2\\cdot\\dfrac{a^2}{2}}{\\dfrac{a\\sqrt{5}}{2}}=\\dfrac{2a}{\\sqrt{5}}$$ +Шаг 6. Находим $AK$. +
В прямоугольном $\\triangle ADK$ (прямой угол при $K$) по теореме Пифагора: +$$AK=\\sqrt{AD^2-DK^2}=\\sqrt{a^2-\\dfrac{4a^2}{5}}=\\sqrt{\\dfrac{a^2}{5}}=\\dfrac{a}{\\sqrt{5}}$$ +Шаг 7. Находим $S_{\\triangle ADK}$. +$$S_{\\triangle ADK}=\\dfrac{1}{2}\\cdot AK\\cdot DK=\\dfrac{1}{2}\\cdot\\dfrac{a}{\\sqrt{5}}\\cdot\\dfrac{2a}{\\sqrt{5}}=\\dfrac{a^2}{5}=4\\text{ см}^2$$ +Шаг 8. Итог. +$$S_{DKMC}=S_{ABCD}-S_{\\triangle ABM}-S_{\\triangle ADK}=20-5-4=11\\text{ см}^2$$ +
Ответ: $11$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v26.js b/frontend/js/exam9/variants/v26.js new file mode 100644 index 0000000..53ce0dc --- /dev/null +++ b/frontend/js/exam9/variants/v26.js @@ -0,0 +1,220 @@ +VARIANTS[26] = { + label: "Вариант 26", + tasks: [ + { + text: `Какое из приведённых ниже выражений тождественно равно произведению $-3(-4-a)$:`, + opts: [ + ["а", "$12-a$"], ["б", "$3a-12$"], ["в", "$12+3a$"], + ["г", "$12-3a$"], ["д", "$3a-4$"], + ], + sol: `Раскрываем скобки: +$$-3(-4-a) = (-3)\\cdot(-4) + (-3)\\cdot(-a) = 12 + 3a$$ +
Ответ: в) $12+3a$
` + }, + { + text: `Значение выражения $126{,}4 - 100{,}97$ равно:`, + opts: [ + ["а", "$25{,}78$"], ["б", "$25{,}52$"], ["в", "$125{,}97$"], + ["г", "$25{,}43$"], ["д", "$125{,}42$"], + ], + sol: `$$126{,}40 - 100{,}97 = 25{,}43$$ +
Ответ: г) $25{,}43$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "если у параллелограмма диагонали перпендикулярны, то это ромб;"], + ["б", "центр вписанной окружности треугольника лежит на пересечении его биссектрис;"], + ["в", "треугольник со сторонами $5$, $12$, $13$ — прямоугольный;"], + ["г", "центральный угол равен половине градусной меры дуги, на которую он опирается?"], + ], + sol: `
    +
  • а) Если диагонали параллелограмма перпендикулярны — это ромб — верно
    • +
  • б) Центр вписанной окружности — точка пересечения биссектрис — верно
    • +
  • в) $5^2+12^2=25+144=169=13^2$ ⟹ прямоугольный — верно
    • +
  • г) «Центральный угол равен половине дуги» — НЕВЕРНО. Центральный угол в градусах равен градусной мере дуги, на которую он опирается (а не половине).
    • +
+
Ответ: г)
` + }, + { + text: `Последовательность $-9;\\; -6;\\; -3;\\; \\ldots$ — арифметическая прогрессия. + Продолжите её далее, записав ещё три члена прогрессии.`, + sol: `Разность прогрессии: $d = -6-(-9) = 3$. +
Продолжаем, прибавляя $3$ к каждому члену: +$$-3+3=0; \\ 0+3=3; \\ 3+3=6$$ +
Ответ: $0;\\;3;\\;6$
` + }, + { + text: `В треугольнике две стороны равны $5$ см и $8$ см, + а внешний угол при вершине, противолежащей третьей стороне, равен $120^{\\circ}$. + Найдите третью сторону треугольника.`, + sol: `Свойство внешнего угла: внешний и внутренний углы при одной вершине в сумме дают $180°$. +
Теорема косинусов: $c^2 = a^2+b^2-2ab\\cos C$, где $C$ — угол между сторонами $a$ и $b$, $c$ — противолежащая сторона. +
Шаг 1. Внешний угол при вершине $C$ равен $120°$. Тогда внутренний угол: +$$\\angle C = 180°-120° = 60°$$ +Шаг 2. По условию вершина $C$ противолежит третьей стороне, значит стороны $a=5$ и $b=8$ выходят из этой вершины — угол $C$ заключён между ними. + + + + C + A + B + 60° + 5 + 8 + $c$ + +Шаг 3. Применяем теорему косинусов, подставив $a=5$, $b=8$, $\\angle C=60°$ и зная, что $\\cos 60°=\\dfrac{1}{2}$: +$$c^2 = a^2+b^2-2ab\\cos C = 5^2+8^2-2\\cdot 5\\cdot 8\\cdot\\dfrac{1}{2}$$ +$$c^2 = 25+64-40 = 49$$ +Шаг 4. Извлекаем корень: +$$c = \\sqrt{49} = 7\\text{ см}$$ +
Ответ: $7$ см
` + }, + { + text: `После проведения профилактических мероприятий необходимо наполнить один из бассейнов + спорткомплекса объёмом $2500$ л. + Через первый кран в бассейн вливается $60$ л воды в минуту. + Сколько литров воды в минуту вливается через второй кран, если при работе двух кранов + одновременно бассейн заполняется через $20$ минут?`, + sol: `Метод введения переменной: неизвестную скорость наполнения второго крана обозначим за $x$. Правило совместной работы: при одновременной работе кранов их производительности складываются. +
Шаг 1. Пусть второй кран вливает $x$ л воды в минуту. +
Шаг 2. Запишем выражение для объёма воды, поступающего в бассейн за $20$ минут. Первый кран за это время даст $20\\cdot 60 = 1200$ л, второй — $20\\cdot x$ л. Суммарно бассейн заполняется на $2500$ л: +$$20\\cdot 60 + 20\\cdot x = 2500.$$ +Шаг 3. Решаем уравнение: +$$1200 + 20x = 2500;$$ +$$20x = 2500-1200 = 1300;$$ +$$x = \\dfrac{1300}{20} = 65\\text{ л/мин}.$$ +
Ответ: $65$ л/мин
` + }, + { + text: `Найдите наименьшее целое число, принадлежащее множеству решений системы неравенств + $$\\begin{cases} \\dfrac{1}{3}(3-6x) + 3x \\leq 1, \\\\[6pt] \\dfrac{1}{2}(2x-12) - 5x \\leq 0. \\end{cases}$$`, + sol: `Метод решения: решаем каждое неравенство отдельно и берём пересечение решений. +
Шаг 1. Решаем первое неравенство. Раскроем скобки: +$$\\dfrac{1}{3}(3-6x)+3x \\leq 1$$ +$$1 - 2x + 3x \\leq 1$$ +$$1 + x \\leq 1 \\;\\implies\\; x \\leq 0$$ +Шаг 2. Решаем второе неравенство. Раскроем скобки: +$$\\dfrac{1}{2}(2x-12) - 5x \\leq 0$$ +$$x - 6 - 5x \\leq 0$$ +$$-4x \\leq 6 \\;\\implies\\; x \\geq -\\dfrac{3}{2}$$ +(при делении на $-4$ знак неравенства меняется) +
Шаг 3. Пересечение решений: $-\\dfrac{3}{2}\\leq x \\leq 0$, то есть $x\\in[-1{,}5;\\;0]$. + + + + −2 + −1 + 0 + + + + −3/2 + +Шаг 4. Целые числа в $[-1{,}5;\\;0]$ — это $-1$ и $0$. Наименьшее из них — $-1$. +
Ответ: $-1$
` + }, + { + text: `Постройте график функции $y = \\dfrac{(x-4)^2}{x-4}$. + Определите, при каких значениях аргумента значение функции не меньше $-2$.`, + sol: `Правило сокращения дроби: множитель из числителя и знаменателя можно сократить только при условии, что он не равен нулю. +
Шаг 1. Найдём ОДЗ. Знаменатель не должен быть равен нулю: +$$x-4\\neq 0 \\;\\implies\\; x\\neq 4.$$ +Шаг 2. Упростим выражение. Числитель $(x-4)^2=(x-4)\\cdot(x-4)$, поэтому при $x\\neq 4$ один множитель $(x-4)$ сокращается: +$$y = \\dfrac{(x-4)^2}{x-4} = x-4,\\quad x\\neq 4.$$ +Значит, график — прямая $y=x-4$ с выколотой точкой при $x=4$, где $y=4-4=0$. + + + + + + + + + + + + + −2 + + 2 + 4 + y=x−4 + y=−2 + x + y + +Шаг 3. Решаем неравенство $y\\geq -2$. Подставляем упрощённое выражение: +$$x-4\\geq -2 \\;\\implies\\; x\\geq 2.$$ +Шаг 4. С учётом ОДЗ исключаем точку $x=4$ из ответа. +
Ответ: $x\\geq 2$, $x\\neq 4$
` + }, + { + text: `На плане размеры прямоугольника $20$ мм $\\times$ $15$ мм. + В реальности площадь прямоугольника равна $300$ см². + Изобразите в заданном масштабе квадрат, если по реальным измерениям + его периметр на $100$ мм больше периметра прямоугольника.`, + sol: `Шаг 1. Масштаб. +
Площадь на плане: $20\\cdot15=300$ мм². В реальности: $300$ см² $=30000$ мм². +$$k^2=\\dfrac{30000}{300}=100 \\implies k=10$$ +Масштаб $1:10$ (1 мм на плане = 10 мм в реальности). +
Шаг 2. Периметр прямоугольника (реальный). +
Реальные размеры: $20\\cdot10=200$ мм и $15\\cdot10=150$ мм. +$$P_{\\text{пр}} = 2(200+150) = 700\\text{ мм}$$ +Шаг 3. Сторона квадрата. +$$P_{\\text{кв}} = 700+100 = 800\\text{ мм} \\implies a = \\dfrac{800}{4}=200\\text{ мм}$$ +Шаг 4. Сторона квадрата на плане. +$$a_{\\text{план}} = \\dfrac{200}{10} = 20\\text{ мм}$$ + + + Прямоугольник + 20 мм × 15 мм + + Квадрат + 20 мм × 20 мм + план (масштаб 1:10) + +
Ответ: квадрат со стороной $20$ мм на плане ($200$ мм в реальности)
` + }, + { + text: `Точка $M$ — середина стороны $BC$ квадрата $ABCD$, площадь которого равна $40$ см². + К отрезку $DM$ проведён перпендикуляр $AK$. + Найдите площадь четырёхугольника $ABMK$.`, + sol: ` + + + + + + + + A + B + C + D + M + K + ABMK + +Формула площади квадрата: $S_{ABCD}=a^2$, где $a$ — сторона. По условию $a^2 = 40$. +
Метод решения: используем координаты, чтобы найти точку $K$ — пересечение прямой $DM$ и проходящего через $A$ перпендикуляра к ней. +
Шаг 1. Введём координаты: $A(0;0)$, $B(a;0)$, $C(a;a)$, $D(0;a)$. Точка $M$ — середина $BC$, поэтому $M\\bigl(a;\\,\\dfrac{a}{2}\\bigr)$. +
Шаг 2. Уравнение прямой $DM$: проходит через $D(0;a)$ и $M(a;\\,a/2)$. Угловой коэффициент: +$$k_{DM} = \\dfrac{a/2-a}{a-0} = -\\dfrac{1}{2}$$ +Уравнение: $y = -\\dfrac{1}{2}x + a$. +
Шаг 3. Прямая $AK$ перпендикулярна $DM$. Произведение угловых коэффициентов перпендикулярных прямых равно $-1$, значит $k_{AK}=2$. Прямая проходит через $A(0;0)$, поэтому $y=2x$. +
Шаг 4. Точка $K$ — пересечение этих прямых: +$$2x = -\\dfrac{1}{2}x + a \\;\\implies\\; \\dfrac{5}{2}x = a \\;\\implies\\; x = \\dfrac{2a}{5}$$ +$$y = 2\\cdot\\dfrac{2a}{5} = \\dfrac{4a}{5}$$ +Значит, $K\\bigl(\\dfrac{2a}{5};\\,\\dfrac{4a}{5}\\bigr)$. +
Шаг 5. Найдём площадь четырёхугольника $ABMK$ (вершины обходятся по порядку $A\\to B\\to M\\to K$) по формуле площади многоугольника через координаты (формула «шнурков»): +$$S = \\dfrac{1}{2}|x_A(y_B-y_K)+x_B(y_M-y_A)+x_M(y_K-y_B)+x_K(y_A-y_M)|$$ +$$= \\dfrac{1}{2}\\left|0+a\\cdot\\dfrac{a}{2}+a\\cdot\\dfrac{4a}{5}+\\dfrac{2a}{5}\\cdot\\left(-\\dfrac{a}{2}\\right)\\right|$$ +$$= \\dfrac{1}{2}\\left|\\dfrac{a^2}{2}+\\dfrac{4a^2}{5}-\\dfrac{a^2}{5}\\right| = \\dfrac{1}{2}\\cdot\\dfrac{11a^2}{10} = \\dfrac{11a^2}{20}$$ +Шаг 6. Подставим $a^2=40$: +$$S = \\dfrac{11\\cdot 40}{20} = 22\\text{ см}^2$$ +
Ответ: $22$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v27.js b/frontend/js/exam9/variants/v27.js new file mode 100644 index 0000000..eb15bde --- /dev/null +++ b/frontend/js/exam9/variants/v27.js @@ -0,0 +1,236 @@ +VARIANTS[27] = { + label: "Вариант 27", + tasks: [ + { + text: `Какое из приведённых ниже выражений тождественно равно произведению $-x(-x+3)$:`, + opts: [ + ["а", "$x^2-3x$"], ["б", "$x^2+3x$"], ["в", "$x^2+3$"], + ["г", "$x^2-3$"], ["д", "$3x-x^2$"], + ], + sol: `Раскрываем скобки: +$$-x(-x+3) = (-x)\\cdot(-x) + (-x)\\cdot3 = x^2 - 3x$$ +
Ответ: а) $x^2-3x$
` + }, + { + text: `Определите промежуток, которому принадлежит значение выражения + $\\left(1\\dfrac{4}{5}+1\\right):2$:`, + opts: [ + ["а", "$(1;\\;2)$"], ["б", "$(2;\\;3)$"], ["в", "$(1;\\;1{,}1)$"], + ["г", "$(0;\\;1)$"], ["д", "$(0;\\;0{,}5)$"], + ], + sol: `Переводим смешанное число: $1\\dfrac{4}{5} = \\dfrac{9}{5}$. +$$\\left(\\dfrac{9}{5}+1\\right):2 = \\dfrac{14}{5}:2 = \\dfrac{14}{10} = \\dfrac{7}{5} = 1{,}4$$ +Так как $1 < 1{,}4 < 2$, значение принадлежит промежутку $(1;\\,2)$. +
Ответ: а) $(1;\\;2)$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "площадь трапеции равна произведению полусуммы оснований на высоту;"], + ["б", "медианы треугольника точкой пересечения делятся в отношении $3:1$;"], + ["в", "касательная к окружности перпендикулярна радиусу, проведённому в точку касания;"], + ["г", "катет в любом прямоугольном треугольнике всегда меньше гипотенузы?"], + ], + sol: `
    +
  • а) $S_{\\text{трап}}=\\dfrac{a+b}{2}\\cdot h$ — верно
    • +
  • б) Медианы делятся точкой пересечения в отношении $\\mathbf{2:1}$ от вершины, а не $3:1$ — НЕВЕРНО
    • +
  • в) Касательная $\\perp$ радиусу в точке касания — верно
    • +
  • г) Катет $<$ гипотенузы — верно
    • +
+
Ответ: б)
` + }, + { + text: `Решите неравенство $\\dfrac{(x-3)^2(x+1)}{(x-2)^3} \\leq 0$ и запишите ответ.`, + sol: `Числитель: $(x-3)^2\\geq0$ всегда, обнуляется при $x=3$. Знак числителя = знак $(x+1)$: +
    +
  • $x < -1$: числитель $< 0$
    • +
  • $x = -1$: числитель $= 0$
    • +
  • $x > -1$, $x\\neq3$: числитель $> 0$
    • +
  • $x = 3$: числитель $= 0$
    • +
+Знаменатель: $(x-2)^3$ имеет знак $(x-2)$: отрицателен при $x<2$, равен $0$ при $x=2$ (ОДЗ: $x\\neq2$), положителен при $x>2$. +
Дробь $\\leq0$: + + + + + + + + + +
СлучайЧислительЗнаменательДробь
$x<-1$$-$$-$$+$ ✗
$x=-1$$0$$-$$0$ ✓
$-1\\lt x\\lt 2$$+$$-$$-$ ✓
$x=2$не определена ✗
$2\\lt x\\lt 3$$+$$+$$+$ ✗
$x=3$$0$$+$$0$ ✓
$x>3$$+$$+$$+$ ✗
+ + + + −1 + 2 + 3 + + + + + ≤ 0 + +
Ответ: $x\\in[-1;\\;2)\\cup\\{3\\}$
` + }, + { + text: `Для украшения двух этажей поместья Деда Мороза было использовано $150$ лампочек. + Для украшения первого этажа потребовалось вдвое больше лампочек, чем для второго. + Сколько лампочек было использовано для украшения второго этажа?`, + sol: `Метод введения переменной: то, о чём спрашивают, обозначим переменной и составим уравнение по условию задачи. +
Шаг 1. Пусть для украшения второго этажа использовали $x$ лампочек. По условию для первого этажа потребовалось вдвое больше, значит $2x$ лампочек. +
Шаг 2. Всего на оба этажа израсходовано $150$ лампочек, значит: +$$x + 2x = 150.$$ +Шаг 3. Решаем уравнение, приводя подобные слагаемые: +$$3x = 150 \\;\\implies\\; x = \\dfrac{150}{3} = 50.$$ +Шаг 4. Значит, для второго этажа использовали $50$ лампочек (а для первого — $2\\cdot 50=100$, сумма $100+50=150$ — сходится с условием). +
Ответ: $50$ лампочек
` + }, + { + text: `Дан треугольник $ABC$, серединные перпендикуляры к его сторонам $AC$ и $BC$ + пересекаются в точке $O$. Докажите, что серединный перпендикуляр к стороне $AB$ + проходит через точку $O$.`, + sol: `Доказательство. + + + + + + + + + + + + + + + + + + + A + B + C + O + OA + OB + OC + ⊥ к AC + ⊥ к BC + ⊥ к AB + OA = OB = OC + +На рисунке: три серединных перпендикуляра (к $AC$, к $BC$, к $AB$) сходятся в одной точке $O$. Двойные засечки показывают $OA=OB=OC$. +

+Точка $O$ лежит на серединном перпендикуляре к $AC$: +$$OA = OC \\quad \\text{(все точки серединного перпендикуляра равноудалены от концов отрезка)}$$ +Точка $O$ лежит на серединном перпендикуляре к $BC$: +$$OB = OC$$ +Из двух равенств: +$$OA = OC = OB \\implies OA = OB$$ +Значит, $O$ равноудалена от $A$ и $B$, то есть лежит на серединном перпендикуляре к $AB$. ∎ +
Доказано: $OA=OC=OB$, поэтому $O$ лежит на серединном перпендикуляре к $AB$
` + }, + { + text: `Решите уравнение $0{,}2(x-2) = 2{,}5 : 0{,}5$ + и запишите число, обратное корню уравнения.`, + sol: `Свойство обратного числа: число, обратное $a$ (при $a\\neq 0$), равно $\\dfrac{1}{a}$. +
Шаг 1. Сначала упростим правую часть. По правилу деления: $2{,}5 : 0{,}5 = \\dfrac{2{,}5}{0{,}5} = 5$. Получаем уравнение: +$$0{,}2(x-2) = 5$$ +Шаг 2. Чтобы найти $x-2$, разделим обе части на $0{,}2$: +$$x-2 = \\dfrac{5}{0{,}2} = 25$$ +Шаг 3. Прибавим $2$ к обеим частям: +$$x = 25 + 2 = 27$$ +Шаг 4. Число, обратное корню $x=27$, равно: +$$\\dfrac{1}{27}$$ +
Ответ: $\\dfrac{1}{27}$
` + }, + { + text: `К задуманному числу $x$, умноженному на $4$, прибавили число, в $2$ раза большее задуманного. + Полученную сумму умножили на $5$ и от полученного произведения вычли число, + в $8$ раз большее $x$. В результате получили число $y$. + Определите вид зависимости числа $y$ от числа $x$.`, + sol: `Прямая пропорциональность: это зависимость вида $y=kx$, где $k$ — постоянное число, отличное от нуля. График такой функции — прямая, проходящая через начало координат. +
Шаг 1. Запишем по условию, что значит «задуманное число $x$, умноженное на $4$»: это $4x$. Число, в $2$ раза большее задуманного, — это $2x$. +
Шаг 2. Найдём сумму этих чисел: +$$4x + 2x = 6x.$$ +Шаг 3. По условию полученную сумму умножили на $5$: +$$5\\cdot 6x = 30x.$$ +Шаг 4. Из этого произведения вычли число, в $8$ раз большее $x$ (то есть $8x$): +$$y = 30x - 8x = 22x.$$ +Шаг 5. Получили зависимость $y=22x$. Так как это запись вида $y=kx$ с $k=22\\neq 0$, перед нами прямая пропорциональность. +
Ответ: прямая пропорциональность $y=22x$
` + }, + { + text: `Сколько решений имеет система уравнений + $$\\begin{cases} x^2 + xy = 15, \\\\[4pt] y^2 + xy = 10? \\end{cases}$$ + Ответ обоснуйте.`, + sol: `Формула квадрата суммы: $(x+y)^2 = x^2+2xy+y^2$. +
Формула разности квадратов: $x^2-y^2 = (x-y)(x+y)$. +
Метод решения: комбинируя уравнения (сложение и вычитание), сводим систему к простой. +
Шаг 1. Сложим уравнения почленно: +$$x^2+xy+y^2+xy = 15+10$$ +$$x^2+2xy+y^2 = 25 \\;\\implies\\; (x+y)^2 = 25$$ +Отсюда $x+y = 5$ или $x+y = -5$. +
Шаг 2. Вычтем второе уравнение из первого: +$$x^2+xy - (y^2+xy) = 15 - 10$$ +$$x^2 - y^2 = 5 \\;\\implies\\; (x-y)(x+y) = 5$$ +Шаг 3. Случай 1: $x+y = 5$. Подставим в $(x-y)(x+y) = 5$: +$$(x-y)\\cdot 5 = 5 \\;\\implies\\; x-y = 1$$ +Решаем систему $x+y=5$ и $x-y=1$: $x=3$, $y=2$. +
Шаг 4. Случай 2: $x+y = -5$. Подставим: +$$(x-y)\\cdot (-5) = 5 \\;\\implies\\; x-y = -1$$ +Решаем систему $x+y=-5$ и $x-y=-1$: $x=-3$, $y=-2$. +
Шаг 5. Проверка. +
Для $(3;\\,2)$: $3^2+3\\cdot 2 = 9+6 = 15$ ✓, $2^2+3\\cdot 2 = 4+6 = 10$ ✓. +
Для $(-3;\\,-2)$: $(-3)^2+(-3)\\cdot(-2) = 9+6 = 15$ ✓, $(-2)^2+(-3)\\cdot(-2) = 4+6 = 10$ ✓. +
Получили ровно $2$ решения. +
Ответ: $2$ решения — $(3;\\,2)$ и $(-3;\\,-2)$
` + }, + { + text: `В параллелограмме $ABCD$ диагонали взаимно перпендикулярны. + Высота $BH$, проведённая к стороне $AD$, пересекает диагональ $AC$ в точке $K$; + $BK = 10$ см, $KH = 6$ см. Найдите площадь параллелограмма.`, + sol: `Шаг 1. Параллелограмм с перпендикулярными диагоналями — ромб. +
В параллелограмме диагонали делятся точкой пересечения пополам. Если они ещё и перпендикулярны, то по теореме Пифагора все стороны равны (ромб). Обозначим сторону $a$, угол $\\angle DAB = \\beta$. + + + + + + + + + A + B + C + D + H + K + O + 10 + 6 + +Шаг 2. Диагональ $AC$ — биссектриса угла $A$ в ромбе. +
В ромбе диагональ делит угол пополам: $\\angle DAC = \\angle BAC = \\dfrac{\\beta}{2}$. +
Шаг 3. Применяем теорему о биссектрисе к $\\triangle ABH$. +
В прямоугольном $\\triangle ABH$ ($\\angle H = 90°$) отрезок $AK$ является биссектрисой угла $A$ (т.к. $AK$ лежит на диагонали ромба $AC$, которая делит $\\angle BAH = \\beta$ пополам). +
По теореме о биссектрисе треугольника биссектриса делит противолежащую сторону в отношении прилежащих сторон: +$$\\dfrac{BK}{KH} = \\dfrac{AB}{AH}$$ +В прямоугольном $\\triangle ABH$: $AH = AB\\cos\\beta$, поэтому: +$$\\dfrac{BK}{KH} = \\dfrac{AB}{AB\\cos\\beta} = \\dfrac{1}{\\cos\\beta}$$ +$$\\cos\\beta = \\dfrac{KH}{BK} = \\dfrac{6}{10} = \\dfrac{3}{5}$$ +Шаг 4. Находим $\\sin\\beta$. +$$\\sin\\beta = \\sqrt{1 - \\cos^2\\beta} = \\sqrt{1 - \\dfrac{9}{25}} = \\sqrt{\\dfrac{16}{25}} = \\dfrac{4}{5}$$ +Шаг 5. Находим сторону ромба $a$. +
В прямоугольном $\\triangle ABH$: $BH = AB\\cdot\\sin\\beta$: +$$BH = BK + KH = 10 + 6 = 16\\text{ см}$$ +$$a = AB = \\dfrac{BH}{\\sin\\beta} = \\dfrac{16}{\\tfrac{4}{5}} = 20\\text{ см}$$ +Шаг 6. Площадь параллелограмма. +$$S = AD\\cdot BH = a\\cdot BH = 20\\cdot16 = 320\\text{ см}^2$$ +
Ответ: $320$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v28.js b/frontend/js/exam9/variants/v28.js new file mode 100644 index 0000000..ac78848 --- /dev/null +++ b/frontend/js/exam9/variants/v28.js @@ -0,0 +1,190 @@ +VARIANTS[28] = { + label: "Вариант 28", + tasks: [ + { + text: `Какое из приведённых ниже выражений тождественно равно произведению $-x(x-4)$:`, + opts: [ + ["а", "$x^2+4x$"], ["б", "$x^2-4x$"], ["в", "$x^2+4$"], + ["г", "$x^2-4$"], ["д", "$4x-x^2$"], + ], + sol: `Раскрываем скобки: +$$-x(x-4) = (-x)\\cdot x + (-x)\\cdot(-4) = -x^2+4x = 4x-x^2$$ +
Ответ: д) $4x-x^2$
` + }, + { + text: `Определите промежуток, которому принадлежит значение выражения + $\\left(1\\dfrac{1}{4}+1\\right):3$:`, + opts: [ + ["а", "$(2;\\;3)$"], ["б", "$(3;\\;4)$"], ["в", "$(0;\\;0{,}5)$"], + ["г", "$(1;\\;2)$"], ["д", "$(0;\\;1)$"], + ], + sol: `Переводим: $1\\dfrac{1}{4} = \\dfrac{5}{4}$. +$$\\left(\\dfrac{5}{4}+1\\right):3 = \\dfrac{9}{4}:3 = \\dfrac{3}{4} = 0{,}75$$ +Так как $0 \\lt 0{,}75 \\lt 1$, значение принадлежит $(0;\\,1)$. +
Ответ: д) $(0;\\;1)$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "площадь трапеции равна произведению средней линии на высоту;"], + ["б", "медианы треугольника точкой пересечения делятся в отношении $1:1$;"], + ["в", "секущая имеет с окружностью ровно две общие точки;"], + ["г", "гипотенуза прямоугольного треугольника всегда больше любого из катетов?"], + ], + sol: `
  • а) верно
  • б) Медианы делятся в отношении $\\mathbf{2:1}$, а не $1:1$ — НЕВЕРНО
  • в) верно
  • г) верно
Ответ: б)
` + }, + { + text: `Решите неравенство $\\dfrac{(x-3)^2(x+1)^2}{(x-2)^4} \\leq 0$ и запишите ответ.`, + sol: `$(x-3)^2(x+1)^2\\geq0$ всегда; $(x-2)^4\\gt0$ при $x\\neq2$. Дробь $\\leq0$ только когда числитель $=0$: $x=3$ или $x=-1$. + + + + −1 + 2 + 3 + + + + = 0 + +
Ответ: $x=-1$ или $x=3$
` + }, + { + text: `В банк заданий для подготовки к экзамену включено $180$ задач. + Из них задач по алгебре на $50$ больше, чем по геометрии. + Сколько задач по алгебре включено в банк?`, + sol: `Метод введения переменной: обозначим за $x$ количество того, чего меньше, и составим уравнение по сумме. +
Шаг 1. Пусть в банке $x$ задач по геометрии. Тогда задач по алгебре на $50$ больше, то есть $x+50$. +
Шаг 2. Всего задач $180$, значит: +$$x + (x+50) = 180.$$ +Шаг 3. Решаем уравнение, приведя подобные слагаемые: +$$2x + 50 = 180 \\;\\implies\\; 2x = 130 \\;\\implies\\; x = 65.$$ +Шаг 4. Это количество задач по геометрии. По алгебре их на $50$ больше: +$$x + 50 = 65 + 50 = 115.$$ +Проверка: $65+115 = 180$ — совпадает с условием. +
Ответ: $115$ задач по алгебре
` + }, + { + text: `Дан треугольник $ABC$, биссектрисы его углов $A$ и $C$ пересекаются в точке $O$. + Докажите, что биссектриса угла $B$ проходит через точку $O$.`, + sol: `Доказательство. + + + + + + + + + + + + + A + B + C + O + бисс. A + бисс. C + бисс. B + d(O,AB)=d(O,BC)=d(O,AC) + +Точка $O$ на биссектрисе угла $A$ — $d(O,AB)=d(O,AC)$. Точка $O$ на биссектрисе угла $C$ — $d(O,BC)=d(O,AC)$. Следовательно $d(O,AB)=d(O,BC)$ — $O$ на биссектрисе угла $B$. ∎
Доказано: $O$ равноудалена от сторон $AB$ и $BC$, значит лежит на биссектрисе угла $B$
` + }, + { + text: `Решите уравнение $(x+2) \\cdot 0{,}5 = \\dfrac{2}{3} \\cdot 2$ + и запишите число, противоположное корню уравнения.`, + sol: `Свойство противоположного числа: число, противоположное $a$, равно $-a$. +
Шаг 1. Вычислим правую часть уравнения: +$$\\dfrac{2}{3}\\cdot 2 = \\dfrac{4}{3}$$ +Получаем уравнение: +$$(x+2)\\cdot 0{,}5 = \\dfrac{4}{3}$$ +Шаг 2. Чтобы найти $x+2$, разделим обе части на $0{,}5$ (то есть умножим на $2$): +$$x+2 = \\dfrac{4}{3}\\cdot 2 = \\dfrac{8}{3}$$ +Шаг 3. Вычтем $2$ из обеих частей: +$$x = \\dfrac{8}{3} - 2 = \\dfrac{8}{3} - \\dfrac{6}{3} = \\dfrac{2}{3}$$ +Шаг 4. Число, противоположное корню $x=\\dfrac{2}{3}$, равно $-\\dfrac{2}{3}$. +
Ответ: $-\\dfrac{2}{3}$
` + }, + { + text: `К задуманному числу $x$, умноженному на $2$, прибавили число, в $3$ раза большее задуманного. + Полученную сумму умножили на $7$ и от полученного произведения вычли число, + в $7$ раз большее $x$. В результате получили число $y$. + Определите вид зависимости числа $y$ от числа $x$.`, + sol: `Прямая пропорциональность: зависимость вида $y=kx$, где $k$ — постоянное число, отличное от нуля. +
Шаг 1. Запишем по условию: задуманное число $x$, умноженное на $2$, — это $2x$. Число, в $3$ раза большее задуманного, — это $3x$. +
Шаг 2. Найдём их сумму: +$$2x + 3x = 5x.$$ +Шаг 3. Полученную сумму умножили на $7$: +$$7\\cdot 5x = 35x.$$ +Шаг 4. Из произведения вычли число, в $7$ раз большее $x$ (то есть $7x$): +$$y = 35x - 7x = 28x.$$ +Шаг 5. Получили зависимость $y=28x$. Это запись вида $y=kx$ с $k=28\\neq 0$, значит перед нами прямая пропорциональность. +
Ответ: прямая пропорциональность $y=28x$
` + }, + { + text: `Сколько решений имеет система уравнений + $$\\begin{cases} x^2 - xy = 9, \\\\[4pt] y^2 - xy = 16? \\end{cases}$$ + Ответ обоснуйте.`, + sol: `Формула квадрата разности: $(x-y)^2 = x^2-2xy+y^2$. +
Формула разности квадратов: $y^2-x^2 = (y-x)(y+x)$. +
Метод решения: комбинируя уравнения (сложение и вычитание), сводим систему к более простой. +
Шаг 1. Сложим уравнения почленно: +$$x^2-xy+y^2-xy = 9+16$$ +$$x^2-2xy+y^2 = 25 \\;\\implies\\; (x-y)^2 = 25$$ +Отсюда $x-y = 5$ или $x-y = -5$. +
Шаг 2. Вычтем первое уравнение из второго: +$$y^2-xy - (x^2-xy) = 16 - 9$$ +$$y^2 - x^2 = 7 \\;\\implies\\; (y-x)(y+x) = 7$$ +Шаг 3. Случай 1: $x-y = 5$, значит $y-x = -5$. Подставляем: +$$-5\\cdot(y+x) = 7 \\;\\implies\\; y+x = -\\dfrac{7}{5}$$ +Решаем систему $x-y=5$ и $x+y=-\\dfrac{7}{5}$. Складывая: $2x = 5-\\dfrac{7}{5} = \\dfrac{18}{5}$, откуда $x=\\dfrac{9}{5}$, а $y = x-5 = -\\dfrac{16}{5}$. +
Шаг 4. Случай 2: $x-y = -5$, значит $y-x = 5$: +$$5\\cdot(y+x) = 7 \\;\\implies\\; y+x = \\dfrac{7}{5}$$ +Решаем $x-y=-5$ и $x+y=\\dfrac{7}{5}$. Складывая: $2x = -5+\\dfrac{7}{5} = -\\dfrac{18}{5}$, поэтому $x=-\\dfrac{9}{5}$, $y=\\dfrac{16}{5}$. +
Шаг 5. Получили ровно $2$ решения: $\\bigl(\\dfrac{9}{5};\\,-\\dfrac{16}{5}\\bigr)$ и $\\bigl(-\\dfrac{9}{5};\\,\\dfrac{16}{5}\\bigr)$. +
Ответ: $2$ решения — $(9/5;\\,-16/5)$ и $(-9/5;\\,16/5)$
` + }, + { + text: `В параллелограмме $ABCD$ диагонали взаимно перпендикулярны. + Высота $BK$, проведённая к стороне $AD$, пересекает диагональ $AC$ в точке $H$; + $HK = 9$ см, $BH = 15$ см. Найдите площадь параллелограмма.`, + sol: ` + + + + + + + + A + B + C + D + K + H + O + 15 + 9 + +Свойство параллелограмма с перпендикулярными диагоналями: такой параллелограмм является ромбом (то есть все стороны равны). +
Свойство диагонали ромба: диагональ ромба делит угол ромба пополам, то есть является биссектрисой. +
Основное тригонометрическое тождество: $\\sin^2\\beta + \\cos^2\\beta = 1$. +
Формула площади параллелограмма: $S = AD\\cdot BK$, где $BK$ — высота к стороне $AD$. +
Шаг 1. Из условия диагонали параллелограмма перпендикулярны, значит $ABCD$ — ромб со стороной $a$. Обозначим $\\angle DAB = \\beta$. +
Шаг 2. Диагональ $AC$ — биссектриса угла $A$, значит $\\angle BAC = \\angle DAC = \\dfrac{\\beta}{2}$. В прямоугольном треугольнике $ABK$ (прямой угол при $K$) точка $H$ лежит на биссектрисе $AH$ угла $A$. +
Шаг 3. По теореме о биссектрисе треугольника $ABK$: +$$\\dfrac{BH}{HK} = \\dfrac{AB}{AK}$$ +В $\\triangle ABK$ имеем $AK = AB\\cos\\beta$. Поэтому: +$$\\dfrac{BH}{HK} = \\dfrac{1}{\\cos\\beta} \\;\\implies\\; \\cos\\beta = \\dfrac{HK}{BH} = \\dfrac{9}{15} = \\dfrac{3}{5}$$ +Шаг 4. По основному тригонометрическому тождеству: +$$\\sin\\beta = \\sqrt{1-\\cos^2\\beta} = \\sqrt{1-\\dfrac{9}{25}} = \\sqrt{\\dfrac{16}{25}} = \\dfrac{4}{5}$$ +Шаг 5. Находим $BK$: это вся высота, $BK = BH + HK = 15 + 9 = 24$ см. +
Шаг 6. Из прямоугольного $\\triangle ABK$: $BK = AB\\cdot\\sin\\beta$, значит: +$$a = AB = \\dfrac{BK}{\\sin\\beta} = \\dfrac{24}{\\tfrac{4}{5}} = 24\\cdot\\dfrac{5}{4} = 30\\text{ см}$$ +Шаг 7. Применяем формулу площади параллелограмма: +$$S = AD\\cdot BK = 30\\cdot 24 = 720\\text{ см}^2$$ +
Ответ: $720$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v29.js b/frontend/js/exam9/variants/v29.js new file mode 100644 index 0000000..912f250 --- /dev/null +++ b/frontend/js/exam9/variants/v29.js @@ -0,0 +1,180 @@ +VARIANTS[29] = { + label: "Вариант 29", + tasks: [ + { + text: `Какой из промежутков является решением неравенства $-x > -3$:`, + opts: [ + ["а", "$(-\\infty;\\; 3)$"], ["б", "$(-\\infty;\\; {-3})$"], ["в", "$(3;\\; {+\\infty})$"], + ["г", "$(-3;\\; {+\\infty})$"], ["д", "$[-3;\\; {+\\infty})$"], + ], + sol: `Делим обе части на $-1$ (знак неравенства меняется): +$$-x > -3 \\implies x < 3$$ + + + + −3 + 0 + 3 + + + $x < 3$ + +
Ответ: а) $(-\\infty;\\;3)$
` + }, + { + text: `Значение выражения $\\dfrac{0{,}6 \\cdot 0{,}3}{-0{,}6}$ равно:`, + opts: [ + ["а", "$0{,}3$"], ["б", "$-0{,}3$"], ["в", "$0{,}18$"], + ["г", "$-0{,}18$"], ["д", "$-0{,}12$"], + ], + sol: `$$\\dfrac{0{,}6\\cdot0{,}3}{-0{,}6} = \\dfrac{0{,}18}{-0{,}6} = -0{,}3$$ +или короче: сокращаем $0{,}6$: $\\dfrac{\\cancel{0{,}6}\\cdot0{,}3}{-\\cancel{0{,}6}} = -0{,}3$. +
Ответ: б) $-0{,}3$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "у любого параллелограмма диагонали перпендикулярны;"], + ["б", "сумма внутренних углов любого четырёхугольника равна $360^{\\circ}$;"], + ["в", "квадрат гипотенузы равен сумме квадратов катетов;"], + ["г", "вписанные углы окружности, опирающиеся на одну и ту же дугу, равны между собой?"], + ], + sol: `
    +
  • а) «У любого параллелограмма диагонали перпендикулярны» — НЕВЕРНО. Диагонали перпендикулярны только в ромбе. Например, в прямоугольнике диагонали равны, но не перпендикулярны.
    • +
  • б) Сумма углов четырёхугольника $= 360°$ — верно
    • +
  • в) Теорема Пифагора — верно
    • +
  • г) Вписанные углы, опирающиеся на одну дугу, равны — верно
    • +
+
Ответ: а)
` + }, + { + text: `Приведите подобные слагаемые $2xy - 4xy + 9x + 4xy - 2x$.`, + sol: `Группируем слагаемые с $xy$ и с $x$: +$$(2xy - 4xy + 4xy) + (9x - 2x) = 2xy + 7x$$ +
Ответ: $2xy + 7x$
` + }, + { + text: `Найдите периметр ромба, диагонали которого равны $18$ см и $24$ см.`, + sol: `Свойство диагоналей ромба: диагонали ромба перпендикулярны и точкой пересечения делятся пополам. +
Теорема Пифагора: в прямоугольном треугольнике $c^2 = a^2 + b^2$. +
Формула периметра ромба: $P = 4a$, где $a$ — сторона. +
Шаг 1. По свойству диагоналей ромба они делятся точкой пересечения пополам и взаимно перпендикулярны. Половины диагоналей: $\\dfrac{18}{2}=9$ см и $\\dfrac{24}{2}=12$ см. + + + + + + + A + B + C + D + O + 9 + 9 + 12 + 12 + a + 18 см + 24 см + +Шаг 2. Половины диагоналей являются катетами прямоугольного треугольника, в котором гипотенуза — сторона ромба $a$. По теореме Пифагора: +$$a = \\sqrt{9^2 + 12^2} = \\sqrt{81+144} = \\sqrt{225} = 15\\text{ см}$$ +Шаг 3. Применяем формулу периметра: +$$P = 4a = 4\\cdot 15 = 60\\text{ см}$$ +
Ответ: $60$ см
` + }, + { + text: `Сократите дробь $\\dfrac{a^2-64}{-a-8}$ и найдите значение полученного выражения при $a = -4$.`, + sol: `Формула разности квадратов: $a^2 - b^2 = (a-b)(a+b)$. +
Шаг 1. Числитель раскладываем по формуле разности квадратов ($64 = 8^2$): +$$a^2-64 = a^2-8^2 = (a-8)(a+8)$$ +Шаг 2. В знаменателе вынесем общий множитель $-1$: +$$-a-8 = -(a+8)$$ +Шаг 3. Сокращаем общий множитель $(a+8)$ (при условии $a\\neq -8$, иначе знаменатель равен нулю): +$$\\dfrac{(a-8)(a+8)}{-(a+8)} = -(a-8) = 8-a$$ +Шаг 4. Подставляем $a = -4$: +$$8-(-4) = 8+4 = 12$$ +
Ответ: $8-a$;  при $a=-4$ значение равно $12$
` + }, + { + text: `Несколько подруг решили обменяться фотографиями на память. + Чтобы каждая девочка получила по одной фотографии от каждой подруги, + потребовалось $42$ фотографии. Сколько было подруг?`, + sol: `Метод введения переменной: неизвестное количество подруг обозначим за $n$. Принцип подсчёта: каждая из $n$ подруг дарит свою фотографию каждой из $n-1$ остальных подруг. +
Шаг 1. Пусть подруг было $n$. Каждая девочка дарит свою фотографию каждой из оставшихся $n-1$ подруг. +
Шаг 2. Общее количество подаренных фотографий — это произведение числа дарителей на число получателей у каждой: +$$n\\cdot(n-1) = 42.$$ +Шаг 3. Раскрываем и приводим к квадратному уравнению: +$$n^2 - n - 42 = 0.$$ +Шаг 4. Решаем через дискриминант: +$$D = (-1)^2 + 4\\cdot 42 = 1+168 = 169 = 13^2;$$ +$$n = \\dfrac{1+13}{2} = 7\\;\\;\\text{(отрицательный корень не подходит, так как $n$ — натуральное)}.$$ +Шаг 5. Проверка: $7\\cdot 6 = 42$ — совпадает с условием. +
Ответ: $7$ подруг
` + }, + { + text: `Найдите $\\text{НОД}(174;\\; 841;\\; 3364)$ и определите, какому множеству он принадлежит: + а) составных чисел; б) простых чисел.`, + sol: `Правило нахождения НОД: наибольший общий делитель — это произведение общих простых множителей чисел, взятых в наименьших степенях. Простое число: натуральное число, которое имеет ровно два различных делителя — $1$ и само число. +
Шаг 1. Разложим каждое число на простые множители. +$$174 = 2\\cdot 87 = 2\\cdot 3\\cdot 29;$$ +$$841 = 29\\cdot 29 = 29^2;$$ +$$3364 = 4\\cdot 841 = 2^2\\cdot 29^2.$$ +Шаг 2. Найдём общие простые множители. В разложении $841$ нет ни $2$, ни $3$, поэтому общий множитель только один — это $29$ (в первой степени, так как наименьшая степень $29$ среди трёх разложений равна $1$): +$$\\text{НОД}(174;\\,841;\\,3364) = 29.$$ +Шаг 3. Определим вид числа $29$. Так как $29$ делится только на $1$ и на $29$, оно является простым. +
Ответ: НОД $= 29$, принадлежит множеству простых чисел (ответ б)
` + }, + { + text: `В арифметической прогрессии сумма трёх первых членов равна $246$. + Чему равна сумма пяти первых членов этой прогрессии, + если её первый член равен разности прогрессии?`, + sol: `Формула $n$-го члена арифметической прогрессии: $a_n = a_1 + (n-1)d$. +
Формула суммы первых $n$ членов арифметической прогрессии: $S_n = \\dfrac{2a_1+(n-1)d}{2}\\cdot n$. +
Шаг 1. По условию $a_1 = d$. Тогда члены прогрессии: +
$a_1 = d$, $a_2 = 2d$, $a_3 = 3d$, $a_4 = 4d$, $a_5 = 5d$. +
Шаг 2. Сумма трёх первых членов: +$$S_3 = a_1+a_2+a_3 = d+2d+3d = 6d$$ +По условию $S_3 = 246$, значит $6d = 246$, откуда $d = 41$. +
Шаг 3. Сумма пяти первых членов: +$$S_5 = d+2d+3d+4d+5d = 15d = 15\\cdot 41 = 615$$ +
Ответ: $S_5 = 615$
` + }, + { + text: `Даны треугольник $ABC$ и окружность, которая проходит через вершины $B$ и $C$ + и пересекает стороны $AB$ и $AC$ соответственно в точках $M$ и $N$, + где $BM = 14$ см, $AN = 8$ см, $NC = 7$ см. + Найдите площадь треугольника $ABC$, если $\\cos A = \\dfrac{\\sqrt{19}}{10}$.`, + sol: ` + + + + + A + B + C + M + N + AM=? + BM=14 + AN=8 + NC=7 + +Шаг 1. Степень точки $A$ относительно окружности. +
Точка $A$ — внешняя, из неё проведены две секущие: $AMB$ и $ANC$. По свойству секущих: +$$AM\\cdot AB = AN\\cdot AC$$ +Обозначим $AM = x$. Тогда $AB = x + 14$, $AC = 8 + 7 = 15$. +$$x(x+14) = 8\\cdot15 = 120$$ +$$x^2 + 14x - 120 = 0$$ +$$D = 196 + 480 = 676 = 26^2$$ +$$x = \\dfrac{-14+26}{2} = 6\\text{ см}$$ +Значит $AM = 6$, $AB = 6+14 = 20$ см. +
Шаг 2. Синус угла $A$. +$$\\sin A = \\sqrt{1-\\cos^2 A} = \\sqrt{1 - \\dfrac{19}{100}} = \\sqrt{\\dfrac{81}{100}} = \\dfrac{9}{10}$$ +Шаг 3. Площадь треугольника $ABC$. +$$S = \\dfrac{1}{2}\\cdot AB\\cdot AC\\cdot\\sin A = \\dfrac{1}{2}\\cdot20\\cdot15\\cdot\\dfrac{9}{10} = \\dfrac{1}{2}\\cdot270 = 135\\text{ см}^2$$ +
Ответ: $135$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v30.js b/frontend/js/exam9/variants/v30.js new file mode 100644 index 0000000..fa4a2b8 --- /dev/null +++ b/frontend/js/exam9/variants/v30.js @@ -0,0 +1,176 @@ +VARIANTS[30] = { + label: "Вариант 30", + tasks: [ + { + text: `Какой из промежутков является решением неравенства $-2x \\geq 1$:`, + opts: [ + ["а", "$(2;\\; {+\\infty})$"], ["б", "$(-2;\\; {+\\infty})$"], ["в", "$(-\\infty;\\; 0{,}5]$"], + ["г", "$(-\\infty;\\; {-0{,}5}]$"], ["д", "$(-\\infty;\\; {-0{,}5})$"], + ], + sol: `Делим обе части на $-2$ (знак неравенства меняется): +$$-2x \\geq 1 \\implies x \\leq -\\dfrac{1}{2} = -0{,}5$$ + + + + 0 + −0,5 + + + $x \\leq -0{,}5$ + +
Ответ: г) $(-\\infty;\\;{-0{,}5}]$
` + }, + { + text: `Значение выражения $\\dfrac{1{,}6 \\cdot 0{,}6}{-0{,}6}$ равно:`, + opts: [ + ["а", "$1{,}6$"], ["б", "$-1{,}6$"], ["в", "$-0{,}6$"], + ["г", "$-0{,}18$"], ["д", "$-0{,}90$"], + ], + sol: `$$\\dfrac{1{,}6\\cdot0{,}6}{-0{,}6} = \\dfrac{0{,}96}{-0{,}6} = -1{,}6$$ +или короче: сокращаем $0{,}6$: $\\dfrac{1{,}6\\cdot\\cancel{0{,}6}}{-\\cancel{0{,}6}} = -1{,}6$. +
Ответ: б) $-1{,}6$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "у любого параллелограмма диагонали равны;"], + ["б", "сумма внутренних углов треугольника равна $180^{\\circ}$;"], + ["в", "если квадрат некоторой стороны треугольника равен сумме квадратов двух других его сторон, то этот треугольник — прямоугольный;"], + ["г", "вписанный угол окружности, опирающийся на диаметр, равен $90^{\\circ}$?"], + ], + sol: `
    +
  • а) «У любого параллелограмма диагонали равны» — НЕВЕРНО. Диагонали равны только у прямоугольника. В произвольном параллелограмме они, как правило, неравны.
    • +
  • б) Сумма углов треугольника равна $180°$ — верно
    • +
  • в) Обратная теорема Пифагора — верно
    • +
  • г) Вписанный угол, опирающийся на диаметр, равен $90°$ — верно
    • +
+
Ответ: а)
` + }, + { + text: `Приведите подобные слагаемые $5xy - 6xy + 9x + xy - 10x$.`, + sol: `Группируем слагаемые с $xy$ и с $x$: +$$(5xy - 6xy + xy) + (9x - 10x) = 0 + (-x) = -x$$ +
Ответ: $-x$
` + }, + { + text: `Найдите периметр ромба, диагонали которого равны $24$ см и $10$ см.`, + sol: `Свойство диагоналей ромба: диагонали ромба перпендикулярны и точкой пересечения делятся пополам. +
Теорема Пифагора: в прямоугольном треугольнике $c^2 = a^2+b^2$. +
Формула периметра ромба: $P = 4a$, где $a$ — сторона. +
Шаг 1. Диагонали ромба делятся точкой пересечения пополам и перпендикулярны. Половины диагоналей: $\\dfrac{24}{2}=12$ см и $\\dfrac{10}{2}=5$ см. + + + + + + + A + B + C + D + O + 12 + 12 + 5 + 5 + a + 24 см + 10 см + +Шаг 2. Половины диагоналей — катеты прямоугольного треугольника, в котором сторона ромба $a$ — гипотенуза. По теореме Пифагора: +$$a = \\sqrt{12^2 + 5^2} = \\sqrt{144+25} = \\sqrt{169} = 13\\text{ см}$$ +Шаг 3. Применяем формулу периметра: +$$P = 4a = 4\\cdot 13 = 52\\text{ см}$$ +
Ответ: $52$ см
` + }, + { + text: `Сократите дробь $\\dfrac{a^2-25}{a+5}$ и найдите значение полученного выражения при $a = -3$.`, + sol: `Формула разности квадратов: $a^2 - b^2 = (a-b)(a+b)$. +
Шаг 1. Числитель раскладываем по формуле разности квадратов ($25 = 5^2$): +$$a^2-25 = a^2-5^2 = (a-5)(a+5)$$ +Шаг 2. Сокращаем общий множитель $(a+5)$ (при условии $a\\neq -5$): +$$\\dfrac{(a-5)(a+5)}{a+5} = a-5$$ +Шаг 3. Подставляем $a = -3$: +$$a-5 = -3-5 = -8$$ +
Ответ: $a-5$;  при $a=-3$ значение равно $-8$
` + }, + { + text: `В соревнованиях по мини-футболу каждая команда сыграла с каждой по одному разу. + Оказалось, что сыграно всего $30$ игр. + Сколько команд участвовало в соревнованиях?`, + sol: `Метод введения переменной: неизвестное количество команд обозначим за $n$ и составим уравнение по подсчёту числа игр. +
Шаг 1. Пусть в соревнованиях участвовало $n$ команд. Каждая из них провела ровно по $n-1$ игр (со всеми остальными командами). +
Шаг 2. Подсчитаем сумму игр всех команд: $n\\cdot(n-1)$. По условию задачи это число равно $30$: +$$n(n-1) = 30.$$ +Шаг 3. Раскроем и получим квадратное уравнение: +$$n^2 - n - 30 = 0.$$ +Шаг 4. Решаем через дискриминант: +$$D = (-1)^2 + 4\\cdot 30 = 1+120 = 121 = 11^2;$$ +$$n = \\dfrac{1+11}{2} = 6\\;\\;\\text{(отрицательный корень не подходит, так как $n$ — натуральное число)}.$$ +Шаг 5. Проверка: $6\\cdot 5 = 30$ — совпадает с условием. +
Ответ: $6$ команд
` + }, + { + text: `Найдите $\\text{НОД}(158;\\; 237;\\; 790)$ и определите, какому множеству он принадлежит: + а) составных чисел; б) простых чисел.`, + sol: `Правило нахождения НОД: наибольший общий делитель — это произведение общих простых множителей чисел, взятых в наименьших степенях. Простое число: натуральное число, имеющее ровно два делителя — $1$ и само число. +
Шаг 1. Разложим каждое число на простые множители: +$$158 = 2\\cdot 79;$$ +$$237 = 3\\cdot 79;$$ +$$790 = 2\\cdot 5\\cdot 79.$$ +Шаг 2. Найдём общие простые множители. В первом числе нет $3$ и $5$, во втором — нет $2$ и $5$. Значит, общим для всех трёх чисел является только множитель $79$: +$$\\text{НОД}(158;\\,237;\\,790) = 79.$$ +Шаг 3. Определим вид числа $79$. Перебором делителей: $79$ не делится на $2,\\,3,\\,5,\\,7$, а $\\sqrt{79}\\lt 9$. Значит, $79$ делится только на $1$ и на $79$ — это простое число. +
Ответ: НОД $= 79$, принадлежит множеству простых чисел (ответ б)
` + }, + { + text: `В арифметической прогрессии сумма четырёх первых членов равна $80$. + Чему равна сумма шести первых членов этой прогрессии, + если её первый член равен разности прогрессии?`, + sol: `Формула $n$-го члена арифметической прогрессии: $a_n = a_1 + (n-1)d$. +
Шаг 1. По условию $a_1 = d$. Тогда члены прогрессии: +
$a_1=d$, $a_2=2d$, $a_3=3d$, $a_4=4d$, $a_5=5d$, $a_6=6d$. +
Шаг 2. Сумма четырёх первых членов: +$$S_4 = d+2d+3d+4d = 10d$$ +По условию $S_4 = 80$, значит $10d = 80$, откуда $d = 8$. +
Шаг 3. Сумма шести первых членов: +$$S_6 = d+2d+3d+4d+5d+6d = 21d = 21\\cdot 8 = 168$$ +
Ответ: $S_6=168$
` + }, + { + text: `Даны треугольник $ABC$ и окружность, которая проходит через вершины $A$ и $B$ + и пересекает стороны $AC$ и $BC$ соответственно в точках $N$ и $M$, + где $AN = 13$ см, $BM = 8$ см, $MC = 4$ см. + Найдите площадь треугольника $ABC$, если $\\cos C = \\dfrac{\\sqrt{11}}{6}$.`, + sol: ` + + + + + C + A + B + N + M + CN=? + AN=13 + CM=4 + MB=8 + +Теорема о двух секущих: если из внешней точки проведены две секущие, то произведения «целая секущая на её внешнюю часть» равны: $CM\\cdot CB = CN\\cdot CA$. +
Основное тригонометрическое тождество: $\\sin^2 C + \\cos^2 C = 1$. +
Формула площади треугольника через две стороны и угол между ними: $S = \\dfrac{1}{2}ab\\sin C$. +
Шаг 1. Точка $C$ — внешняя для окружности. Через неё проходят две секущие: $CMB$ и $CNA$, причём $CB = CM+MB = 4+8 = 12$ см. Обозначим $CN = x$, тогда $CA = x+13$ см. +
Шаг 2. По теореме о двух секущих: +$$CM\\cdot CB = CN\\cdot CA$$ +$$4\\cdot 12 = x(x+13)$$ +$$x^2 + 13x - 48 = 0$$ +По теореме Виета или через дискриминант: $D = 169 + 192 = 361 = 19^2$, $x = \\dfrac{-13+19}{2} = 3$ (берём положительный корень). Значит $CN = 3$, $CA = 16$ см. +
Шаг 3. Находим $\\sin C$ по основному тригонометрическому тождеству: +$$\\sin C = \\sqrt{1-\\cos^2 C} = \\sqrt{1-\\dfrac{11}{36}} = \\sqrt{\\dfrac{25}{36}} = \\dfrac{5}{6}$$ +Шаг 4. Площадь треугольника $ABC$ по формуле через две стороны и угол $C$ (это угол между сторонами $CA$ и $CB$): +$$S = \\dfrac{1}{2}\\cdot CA\\cdot CB\\cdot\\sin C = \\dfrac{1}{2}\\cdot 16\\cdot 12\\cdot\\dfrac{5}{6} = 80\\text{ см}^2$$ +
Ответ: $80$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v31.js b/frontend/js/exam9/variants/v31.js new file mode 100644 index 0000000..62c2b0e --- /dev/null +++ b/frontend/js/exam9/variants/v31.js @@ -0,0 +1,209 @@ +VARIANTS[31] = { + label: "Вариант 31", + tasks: [ + { + text: `Значение выражения $3 : \\dfrac{3}{5} + 7$ равно:`, + opts: [ + ["а", "$9$"], ["б", "$12$"], ["в", "$8{,}4$"], + ["г", "$7{,}2$"], ["д", "$11$"], + ], + sol: `$$3:\\dfrac{3}{5}+7 = 3\\cdot\\dfrac{5}{3}+7 = 5+7 = 12$$ +
Ответ: б) $12$
` + }, + { + text: `Запись выражения $\\dfrac{3m}{n^3} \\cdot \\dfrac{n}{m}$ в виде дроби имеет вид:`, + opts: [ + ["а", "$\\dfrac{3}{n^3}$"], ["б", "$\\dfrac{n^4}{3m^2}$"], ["в", "$\\dfrac{3}{n^2}$"], + ["г", "$\\dfrac{3m^2}{n^4}$"], ["д", "$\\dfrac{3n}{m}$"], + ], + sol: `Сокращаем $m$ и одну степень $n$: +$$\\dfrac{3m}{n^3}\\cdot\\dfrac{n}{m} = \\dfrac{3\\cancel{m}\\cdot\\cancel{n}}{n^3\\cdot\\cancel{m}\\cdot\\cancel{1}} = \\dfrac{3}{n^2}$$ +
Ответ: в) $\\dfrac{3}{n^2}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "окружность, вписанная в треугольник, касается всех его сторон;"], + ["б", "синус острого угла прямоугольного треугольника равен отношению прилежащего катета к гипотенузе;"], + ["в", "средняя линия треугольника равна половине его основания;"], + ["г", "радиус описанной окружности треугольника со стороной $a$ и углом $\\alpha$ находится из формулы $\\dfrac{a}{\\sin\\alpha} = 2R$?"], + ], + sol: `
    +
  • а) Вписанная окружность касается всех сторон — верно
    • +
  • б) «Синус равен отношению прилежащего катета к гипотенузе» — НЕВЕРНО. Синус = противолежащий катет / гипотенуза. Прилежащий катет / гипотенуза — это косинус.
    • +
  • в) Средняя линия треугольника $=$ половина основания — верно
    • +
  • г) Теорема синусов: $\\dfrac{a}{\\sin\\alpha}=2R$ — верно
    • +
+
Ответ: б)
` + }, + { + text: `Расстояние между городами на карте $5$ см. + Определите это расстояние на местности, если масштаб карты $1 : 100\\,000$.`, + sol: `Масштаб $1:100\\,000$ означает: $1$ см на карте $= 100\\,000$ см $= 1$ км на местности. +$$5\\text{ см}\\times100\\,000 = 500\\,000\\text{ см} = 5\\text{ км}$$ +
Ответ: $5$ км
` + }, + { + text: `На покраску пола в спортивном зале школы израсходовали $32$ кг краски, + что составило $\\dfrac{1}{4}$ всей купленной краски. + Сколько всего килограммов краски было куплено?`, + sol: `Правило нахождения числа по его части: чтобы найти всё число, зная его часть $\\dfrac{m}{n}$, нужно эту часть разделить на $\\dfrac{m}{n}$ (или умножить на обратную дробь $\\dfrac{n}{m}$). +
Шаг 1. Пусть всего было куплено $x$ кг краски. По условию израсходованные $32$ кг составляют $\\dfrac{1}{4}$ от $x$: +$$\\dfrac{1}{4}\\cdot x = 32.$$ +Шаг 2. Умножим обе части уравнения на $4$, чтобы найти $x$: +$$x = 32\\cdot 4 = 128\\text{ кг}.$$ +Проверка: $\\dfrac{1}{4}\\cdot 128 = 32$ — совпадает с условием. +
Ответ: $128$ кг
` + }, + { + text: `Найдите наименьшее целое решение двойного неравенства $-12 < 2x - 6 \\leq 4$.`, + sol: `Метод решения двойного неравенства: выполняем одинаковые действия со всеми тремя частями (прибавлять, вычитать, умножать или делить на положительное число — знак сохраняется; на отрицательное — меняется). +
Шаг 1. Прибавим $6$ ко всем трём частям: +$$-12+6 \\lt 2x-6+6 \\leq 4+6$$ +$$-6 \\lt 2x \\leq 10$$ +Шаг 2. Разделим все части на $2$ (положительное число, знак не меняется): +$$-3 \\lt x \\leq 5$$ + + + + −3 + −2 + 0 + 5 + + + + + +Шаг 3. Так как левое неравенство строгое, $x=-3$ не подходит. Наименьшее целое число, строго большее $-3$, — это $-2$. +
Ответ: $-2$
` + }, + { + text: `$ABCD$ — прямоугольник с периметром $28$ см, у которого $AC = 10$ см. + Найдите радиус окружности, вписанной в треугольник $ABD$.`, + sol: `Шаг 1. Находим стороны прямоугольника. +
Пусть $AB = a$, $AD = b$. Тогда: +$$2(a+b)=28 \\implies a+b=14$$ +$$a^2+b^2 = AC^2 = 100$$ +$$(a+b)^2 = a^2+2ab+b^2 \\implies 196 = 100+2ab \\implies ab=48$$ +Из $a+b=14$, $ab=48$: решаем $t^2-14t+48=0 \\implies t=6$ или $t=8$. +Стороны: $AB=8$, $AD=6$ (или наоборот). + + + + + + + A + B + C + D + 8 см + 6 + r + BD=10 + +Шаг 2. Треугольник $ABD$. +
В прямоугольнике $\\angle DAB=90°$, поэтому $\\triangle ABD$ — прямоугольный с прямым углом при $A$. +Катеты: $AB=8$, $AD=6$. Гипотенуза: $BD=AC=10$. +
Шаг 3. Радиус вписанной окружности прямоугольного треугольника: +$$r = \\dfrac{AB + AD - BD}{2} = \\dfrac{8+6-10}{2} = \\dfrac{4}{2} = 2\\text{ см}$$ +
Ответ: $r = 2$ см
` + }, + { + text: `При каких действительных значениях $a$ график функции $y = x^2 - 5x + 5a$ + имеет с осью абсцисс единственную общую точку?`, + sol: `Условие единственной общей точки параболы с осью $Ox$: уравнение $y=0$ должно иметь ровно один корень. Для квадратного уравнения $Ax^2+Bx+C=0$ это значит, что дискриминант $D = B^2-4AC$ равен нулю. +
Шаг 1. Точки пересечения графика с осью $Ox$ — это корни уравнения $y=0$, то есть: +$$x^2 - 5x + 5a = 0.$$ +Шаг 2. Чтобы было ровно одно решение, нужно $D=0$. Вычислим дискриминант ($A=1$, $B=-5$, $C=5a$): +$$D = (-5)^2 - 4\\cdot 1\\cdot 5a = 25 - 20a.$$ +Шаг 3. Приравниваем дискриминант к нулю и решаем: +$$25 - 20a = 0 \\;\\implies\\; 20a = 25 \\;\\implies\\; a = \\dfrac{25}{20} = \\dfrac{5}{4}.$$ +
Ответ: $a = \\dfrac{5}{4}$
` + }, + { + text: `Какое наименьшее число членов прогрессии $32{,}5;\\; 37{,}5;\\; 42{,}5;\\; \\ldots$ + нужно взять, чтобы их сумма была больше $2160$?`, + sol: `Формула суммы первых $n$ членов арифметической прогрессии: $S_n = \\dfrac{2a_1+(n-1)d}{2}\\cdot n$. +
Шаг 1. Из условия $a_1=32{,}5$. Разность прогрессии $d = 37{,}5-32{,}5 = 5$. +
Шаг 2. Запишем формулу суммы: +$$S_n = \\dfrac{2\\cdot 32{,}5+(n-1)\\cdot 5}{2}\\cdot n = \\dfrac{65+5n-5}{2}\\cdot n = \\dfrac{n(60+5n)}{2}$$ +Шаг 3. Запишем неравенство $S_n \\gt 2160$: +$$\\dfrac{n(60+5n)}{2} \\gt 2160$$ +$$n(60+5n) \\gt 4320$$ +$$5n^2+60n - 4320 \\gt 0$$ +$$n^2+12n-864 \\gt 0$$ +Шаг 4. Решаем квадратное уравнение $n^2+12n-864 = 0$: +$$D = 12^2+4\\cdot 864 = 144+3456 = 3600$$ +$$\\sqrt{D} = 60,\\quad n = \\dfrac{-12+60}{2} = 24$$ +Неравенство выполняется при $n \\gt 24$ (поскольку коэффициент при $n^2$ положителен и нас интересуют натуральные $n$). +
Шаг 5. Проверим: при $n=24$ имеем $S_{24} = \\dfrac{24\\cdot(60+120)}{2} = \\dfrac{24\\cdot 180}{2} = 2160$ — равно, но нужно строго больше. +
При $n=25$: $S_{25} = \\dfrac{25\\cdot(60+125)}{2} = \\dfrac{25\\cdot 185}{2} = 2312{,}5 \\gt 2160$ ✓. +
Значит, наименьшее $n = 25$. +
Ответ: наименьшее число членов $= 25$
` + }, + { + text: `Дана трапеция $ABCD$ с основаниями $AD$ и $BC$, $AB = CD$, + диагональ $AC$ перпендикулярна стороне $CD$, угол $BAC$ равен углу $DAC$. + Найдите площадь трапеции, если площадь треугольника $ACD$ равна $6$ см².`, + sol: `Шаг 1. Находим угол при вершине $A$. +
Пусть $\\angle DAC = \\angle BAC = \\alpha$. В прямоугольном $\\triangle ACD$ ($\\angle ACD=90°$): +$$\\angle DAC + \\angle ADC = 90° \\implies \\angle ADC = 90°-\\alpha$$ +В равнобедренной трапеции ($AB=CD$) углы при основании $AD$ равны: $\\angle ADC = \\angle DAB$. +$$\\angle DAB = 2\\alpha \\implies 2\\alpha = 90°-\\alpha \\implies \\alpha = 30°$$ +Значит, $\\angle DAB = \\angle ADC = 60°$. +
Шаг 2. Определяем размеры трапеции. +
Треугольник $ACD$: $\\angle ACD=90°$, $\\angle DAC=30°$, $\\angle ADC=60°$ — это прямоугольный $30°$-$60°$-$90°$. +
Пусть $CD = a$. Тогда $AC = a\\sqrt{3}$, $AD = 2a$. + + + + + + + + + + + + + + + + + + + + + A + B + C + D + + 30° + 30° + 60° + + 90° + + 6 см² + 3 см² + + a + a + a + 2a + +Шаг 3. Треугольник $ABC$. +
В трапеции $AD\\|BC$, поэтому $\\angle DAB + \\angle ABC = 180°$, откуда $\\angle ABC = 120°$. +
В $\\triangle ABC$: $\\angle BAC=30°$, $\\angle ABC=120°$, значит $\\angle BCA = 180°-30°-120°=30°$. +
Поскольку $\\angle BAC = \\angle BCA = 30°$, треугольник $ABC$ — равнобедренный: $AB = BC = a$. +
Шаг 4. Площади. +$$S_{ACD} = \\tfrac{1}{2}\\cdot AC\\cdot CD = \\tfrac{1}{2}\\cdot a\\sqrt{3}\\cdot a = \\dfrac{a^2\\sqrt{3}}{2} = 6$$ +$$S_{ABC} = \\tfrac{1}{2}\\cdot AB\\cdot BC\\cdot\\sin(\\angle ABC) = \\tfrac{1}{2}\\cdot a\\cdot a\\cdot\\sin120° = \\dfrac{a^2\\sqrt{3}}{4} = \\dfrac{1}{2}\\cdot6 = 3\\text{ см}^2$$ +$$S_{ABCD} = S_{ACD} + S_{ABC} = 6 + 3 = 9\\text{ см}^2$$ +
Ответ: $9$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v32.js b/frontend/js/exam9/variants/v32.js new file mode 100644 index 0000000..2ff0a3b --- /dev/null +++ b/frontend/js/exam9/variants/v32.js @@ -0,0 +1,189 @@ +VARIANTS[32] = { + label: "Вариант 32", + tasks: [ + { + text: `Значение выражения $7 : \\dfrac{7}{9} - 5$ равно:`, + opts: [ + ["а", "$1$"], ["б", "$14$"], ["в", "$4$"], + ["г", "$3$"], ["д", "$2$"], + ], + sol: `$7:\\dfrac{7}{9}-5 = 7\\cdot\\dfrac{9}{7}-5 = 9-5 = 4$.
Ответ: в) $4$
` + }, + { + text: `Запись выражения $\\dfrac{5a}{b^4} \\cdot \\dfrac{b}{a}$ в виде дроби имеет вид:`, + opts: [ + ["а", "$\\dfrac{5}{b^4}$"], ["б", "$\\dfrac{5a^2}{b^5}$"], ["в", "$\\dfrac{5}{b^3}$"], + ["г", "$\\dfrac{b^5}{5a^2}$"], ["д", "$\\dfrac{b}{a}$"], + ], + sol: `Сокращаем $a$ и $b$: $\\dfrac{5\\cancel{a}}{b^4}\\cdot\\dfrac{\\cancel{b}}{\\cancel{a}}=\\dfrac{5}{b^3}$.
Ответ: в) $\\dfrac{5}{b^3}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "окружность, описанная около треугольника, проходит через все его вершины;"], + ["б", "косинус острого угла прямоугольного треугольника равен отношению противолежащего катета к гипотенузе;"], + ["в", "средняя линия трапеции равна полусумме её оснований;"], + ["г", "радиус окружности, вписанной в треугольник, находится из формулы $S = pr$?"], + ], + sol: `а) Описанная окружность проходит через все вершины — верно. б) «Косинус = противолежащий катет / гипотенуза» — НЕВЕРНО: это синус. Косинус = прилежащий / гипотенуза. в) Средняя линия трапеции = полусумма оснований — верно. г) $S=pr$ — верно.
Ответ: б)
` + }, + { + text: `Расстояние между городами на карте $9$ см. + Определите это расстояние на местности, если масштаб карты $1 : 1\\,000\\,000$.`, + sol: `Масштаб $1:1\\,000\\,000$: $9$ см $\\times 1\\,000\\,000 = 9\\,000\\,000$ см $= 90$ км.
Ответ: $90$ км
` + }, + { + text: `На подкормку рассады овощей в теплице израсходовали $12$ кг удобрений, + что составило $\\dfrac{1}{6}$ всей массы удобрений, купленных фермером. + Сколько всего килограммов удобрений было куплено?`, + sol: `Правило нахождения числа по его части: чтобы найти всё число, зная его часть $\\dfrac{m}{n}$, нужно эту часть разделить на $\\dfrac{m}{n}$ (или умножить на обратную дробь $\\dfrac{n}{m}$). +
Шаг 1. Пусть всего было куплено $x$ кг удобрений. По условию израсходованные $12$ кг составляют $\\dfrac{1}{6}$ от $x$: +$$\\dfrac{1}{6}\\cdot x = 12.$$ +Шаг 2. Умножим обе части уравнения на $6$, чтобы выразить $x$: +$$x = 12\\cdot 6 = 72\\text{ кг}.$$ +Проверка: $\\dfrac{1}{6}\\cdot 72 = 12$ — совпадает с условием. +
Ответ: $72$ кг
` + }, + { + text: `Найдите наибольшее целое решение двойного неравенства $-9 \\leq 3x - 6 < 6$.`, + sol: `Метод решения двойного неравенства: выполняем одинаковые действия со всеми тремя частями. При умножении или делении на отрицательное число знаки неравенств меняются; на положительное — сохраняются. +
Шаг 1. Прибавим $6$ ко всем трём частям, чтобы избавиться от $-6$ в средней части: +$$-9+6 \\leq 3x-6+6 \\lt 6+6$$ +$$-3 \\leq 3x \\lt 12$$ +Шаг 2. Разделим все части на $3$ (положительное число, знаки не меняются): +$$-1 \\leq x \\lt 4$$ + + + + −1 + 0 + 3 + 4 + + + + + +Шаг 3. Правое неравенство строгое, поэтому $x=4$ не подходит. Наибольшее целое, строго меньшее $4$, — это $3$. +
Ответ: $3$
` + }, + { + text: `$ABCD$ — прямоугольник с периметром $42$ см, у которого $BD = 15$ см. + Найдите радиус окружности, вписанной в треугольник $ADC$.`, + sol: `Формула периметра прямоугольника: $P = 2(a+b)$. +
Теорема Пифагора: в прямоугольном треугольнике $c^2 = a^2+b^2$. В прямоугольнике диагональ образует с двумя сторонами прямоугольный треугольник. +
Формула квадрата суммы: $(a+b)^2 = a^2+2ab+b^2$. +
Формула радиуса вписанной окружности прямоугольного треугольника: $r = \\dfrac{a+b-c}{2}$, где $a$, $b$ — катеты, $c$ — гипотенуза. +
Шаг 1. Обозначим стороны прямоугольника $AB = a$, $AD = b$. Из периметра: +$$2(a+b) = 42 \\;\\implies\\; a+b = 21$$ +Шаг 2. Диагональ $BD$ — гипотенуза прямоугольного $\\triangle ABD$ (прямой угол при $A$). По теореме Пифагора: +$$a^2+b^2 = BD^2 = 15^2 = 225$$ +Шаг 3. Найдём $ab$ через квадрат суммы: +$$(a+b)^2 = a^2+2ab+b^2 \\;\\implies\\; 21^2 = 225+2ab$$ +$$441 = 225+2ab \\;\\implies\\; 2ab = 216 \\;\\implies\\; ab = 108$$ +Шаг 4. Стороны $a$ и $b$ — корни уравнения $t^2-21t+108=0$: +$$D = 21^2-4\\cdot 108 = 441-432 = 9, \\quad \\sqrt{D} = 3$$ +$$t_{1,2} = \\dfrac{21\\pm 3}{2} = 12\\text{ или } 9$$ +Значит $AB=12$ см, $AD=9$ см. +
Шаг 5. Треугольник $ADC$ — прямоугольный с прямым углом при $D$ (стороны прямоугольника перпендикулярны). Катеты: $AD=9$, $DC=AB=12$. Гипотенуза $AC$ равна $BD=15$ (диагонали прямоугольника равны). + + + + + + + A + B + C + D + 12 см + 9 + r + AC=15 + +Шаг 6. Применяем формулу радиуса вписанной окружности прямоугольного треугольника: +$$r = \\dfrac{AD+DC-AC}{2} = \\dfrac{9+12-15}{2} = \\dfrac{6}{2} = 3\\text{ см}$$ +
Ответ: $r=3$ см
` + }, + { + text: `При каких действительных значениях $a$ график функции $y = x^2 - 6x + 3a$ + имеет с осью абсцисс единственную общую точку?`, + sol: `Условие единственной общей точки параболы с осью $Ox$: уравнение $y=0$ должно иметь ровно один корень. Для квадратного уравнения $Ax^2+Bx+C=0$ это значит, что дискриминант $D=B^2-4AC$ равен нулю. +
Шаг 1. Точки пересечения с осью $Ox$ — это корни уравнения $y=0$: +$$x^2 - 6x + 3a = 0.$$ +Шаг 2. Чтобы было ровно одно решение, нужно $D=0$. Вычислим дискриминант ($A=1$, $B=-6$, $C=3a$): +$$D = (-6)^2 - 4\\cdot 1\\cdot 3a = 36 - 12a.$$ +Шаг 3. Приравниваем к нулю и решаем: +$$36 - 12a = 0 \\;\\implies\\; 12a = 36 \\;\\implies\\; a = 3.$$ +
Ответ: $a = 3$
` + }, + { + text: `Какое наименьшее число членов прогрессии $31{,}5;\\; 36{,}5;\\; 41{,}5;\\; \\ldots$ + нужно взять, чтобы их сумма была больше $84$?`, + sol: `Формула суммы первых $n$ членов арифметической прогрессии: $S_n = \\dfrac{2a_1+(n-1)d}{2}\\cdot n$. +
Шаг 1. Из условия $a_1 = 31{,}5$. Разность прогрессии $d = 36{,}5-31{,}5 = 5$. +
Шаг 2. Запишем формулу суммы: +$$S_n = \\dfrac{2\\cdot 31{,}5 + (n-1)\\cdot 5}{2}\\cdot n = \\dfrac{63+5n-5}{2}\\cdot n = \\dfrac{n(58+5n)}{2}$$ +Шаг 3. Условие $S_n \\gt 84$: +$$\\dfrac{n(58+5n)}{2} \\gt 84$$ +$$5n^2+58n - 168 \\gt 0$$ +Шаг 4. Решаем уравнение $5n^2+58n-168=0$: +$$D = 58^2+4\\cdot 5\\cdot 168 = 3364+3360 = 6724 = 82^2$$ +$$n = \\dfrac{-58+82}{10} = \\dfrac{24}{10} = 2{,}4$$ +Неравенство выполняется при $n \\gt 2{,}4$ (так как коэффициент при $n^2$ положителен). +
Шаг 5. Наименьшее натуральное $n$, удовлетворяющее $n\\gt 2{,}4$, — это $n=3$. +
Проверка: $S_3 = 31{,}5+36{,}5+41{,}5 = 109{,}5 \\gt 84$ ✓. +
Ответ: $3$
` + }, + { + text: `Дана трапеция $ABCD$ с основаниями $AD$ и $BC$, $AB = CD$, + диагональ $AC$ перпендикулярна стороне $CD$, угол $BAC$ равен углу $DAC$. + Найдите площадь трапеции, если площадь треугольника $ADC$ равна $12$ см².`, + sol: ` + + + + + + + + + + + A + B + C + D + 30° + 30° + 60° + 90° + 12 см² + 6 см² + a + a + a + 2a + +Свойство равнобедренной трапеции: в равнобедренной трапеции углы при основании равны. +
Сумма углов при боковой стороне трапеции: равна $180°$, так как основания параллельны. +
Свойство прямоугольного треугольника $30°$-$60°$-$90°$: катет, лежащий против угла $30°$, равен половине гипотенузы. +
Шаг 1. Обозначим $\\angle DAC = \\angle BAC = \\alpha$. Тогда $\\angle DAB = 2\\alpha$. +
Шаг 2. В прямоугольном $\\triangle ACD$ ($\\angle ACD = 90°$ по условию): +$$\\angle ADC = 90°-\\alpha$$ +Шаг 3. Поскольку $ABCD$ — равнобедренная трапеция ($AB=CD$), углы при большем основании равны: $\\angle DAB = \\angle ADC$, то есть $2\\alpha = 90°-\\alpha$. Отсюда $\\alpha=30°$, значит $\\angle DAC=30°$, $\\angle ADC=60°$. +
Шаг 4. В $\\triangle ACD$ обозначим $CD=a$. По свойству прямоугольного треугольника $30°$-$60°$-$90°$ катет $CD$ напротив $30°$ равен половине гипотенузы $AD$, поэтому $AD=2a$. Тогда $AC=a\\sqrt{3}$ (по теореме Пифагора или свойству). +
Шаг 5. Найдём углы $\\triangle ABC$. Так как $AD\\|BC$, имеем $\\angle DAB+\\angle ABC=180°$, поэтому $\\angle ABC=180°-60°=120°$. Из суммы углов $\\triangle ABC$: $\\angle BCA = 180°-30°-120°=30°$. Поскольку $\\angle BAC=\\angle BCA=30°$, треугольник $ABC$ равнобедренный: $AB=BC=a$. +
Шаг 6. Площадь треугольника $ABC$ (через две стороны и угол между ними): +$$S_{ABC} = \\dfrac{1}{2}\\cdot AB\\cdot BC\\cdot\\sin\\angle ABC = \\dfrac{1}{2}\\cdot a\\cdot a\\cdot\\sin 120° = \\dfrac{a^2\\sqrt{3}}{4}$$ +А площадь $\\triangle ACD$ (прямоугольный, катеты $a$ и $a\\sqrt{3}$): +$$S_{ACD} = \\dfrac{1}{2}\\cdot a\\cdot a\\sqrt{3} = \\dfrac{a^2\\sqrt{3}}{2}$$ +Видно, что $S_{ABC} = \\dfrac{1}{2}\\cdot S_{ACD}$. +
Шаг 7. По условию $S_{ACD}=12$ см², значит: +$$S_{ABC} = \\dfrac{1}{2}\\cdot 12 = 6\\text{ см}^2$$ +Шаг 8. Площадь всей трапеции: +$$S_{ABCD} = S_{ACD} + S_{ABC} = 12 + 6 = 18\\text{ см}^2$$ +
Ответ: $18$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v33.js b/frontend/js/exam9/variants/v33.js new file mode 100644 index 0000000..f837b37 --- /dev/null +++ b/frontend/js/exam9/variants/v33.js @@ -0,0 +1,194 @@ +VARIANTS[33] = { + label: "Вариант 33", + tasks: [ + { + text: `Отношение чисел $16$ и $2$ равно:`, + opts: [ + ["а", "$8$"], ["б", "$32$"], ["в", "$16$"], + ["г", "$14$"], ["д", "$1$"], + ], + sol: `$$16 : 2 = 8$$ +
Ответ: а) $8$
` + }, + { + text: `Запись выражения $\\dfrac{m}{n} - 1$ в виде дроби имеет вид:`, + opts: [ + ["а", "$\\dfrac{m-1}{n}$"], ["б", "$\\dfrac{1-m}{n}$"], ["в", "$\\dfrac{m-n}{n}$"], + ["г", "$m - n$"], ["д", "$\\dfrac{n-m}{n}$"], + ], + sol: `Приводим к общему знаменателю $n$: +$$\\dfrac{m}{n} - 1 = \\dfrac{m}{n} - \\dfrac{n}{n} = \\dfrac{m-n}{n}$$ +
Ответ: в) $\\dfrac{m-n}{n}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "если четырёхугольник $ABCD$ вписан в окружность, то $\\angle A + \\angle C = 180^{\\circ}$;"], + ["б", "любая медиана равнобедренного треугольника является его высотой;"], + ["в", "вписанный угол равен половине градусной меры дуги, на которую он опирается;"], + ["г", "тангенс острого угла прямоугольного треугольника равен отношению противолежащего катета к прилежащему?"], + ], + sol: `
    +
  • а) Противоположные углы вписанного четырёхугольника дополняют друг друга до $180°$ — верно
    • +
  • б) «Любая медиана равнобедренного треугольника — это высота» — НЕВЕРНО. Медианой-высотой является только медиана, проведённая из вершинного угла к основанию. Медианы к боковым сторонам высотами не являются.
    • +
  • в) Вписанный угол $= \\frac{1}{2}$ дуги — верно
    • +
  • г) $\\tan\\alpha = \\dfrac{\\text{противолежащий катет}}{\\text{прилежащий катет}}$ — верно
    • +
+
Ответ: б)
` + }, + { + text: `Решите неравенство $\\dfrac{x}{2} + 3 > 0$ и укажите наименьшее целое решение этого неравенства.`, + sol: `$$\\dfrac{x}{2} > -3 \\implies x > -6$$ +Решение: $x\\in(-6;\\,+\\infty)$. Наименьшее целое число, строго большее $-6$ — это $\\mathbf{-5}$. +
Ответ: $x > -6$; наименьшее целое решение $= -5$
` + }, + { + text: `В угол $A$ вписана окружность с центром в точке $O$, которая касается сторон угла + в точках $B$ и $C$. Найдите угол $BCO$, если $\\angle A = 64^{\\circ}$.`, + sol: `Свойство касательной к окружности: радиус, проведённый в точку касания, перпендикулярен касательной. +
Значит, $\\angle OBA = 90°$ и $\\angle OCA = 90°$. + + + + + + + + + + + + + + A + O + B + C + 64° + 32° + ? + +Шаг 1. Найдём угол $BOC$ в четырёхугольнике $ABOC$. +
По теореме о сумме углов четырёхугольника: сумма углов любого четырёхугольника равна $360°$. +$$\\angle A + \\angle OBA + \\angle BOC + \\angle OCA = 360°$$ +Подставляем известные углы: +$$64° + 90° + \\angle BOC + 90° = 360°$$ +$$\\angle BOC = 360° - 244° = 116°$$ +Шаг 2. Рассмотрим треугольник $OBC$. +
Так как $OB$ и $OC$ — радиусы одной окружности, то $OB = OC$. Значит, $\\triangle OBC$ равнобедренный с основанием $BC$. +
По свойству равнобедренного треугольника: углы при основании равны. +$$\\angle OBC = \\angle BCO$$ +Шаг 3. Сумма углов треугольника $OBC$ равна $180°$: +$$\\angle BCO = \\dfrac{180° - \\angle BOC}{2} = \\dfrac{180° - 116°}{2} = \\dfrac{64°}{2} = 32°$$ +
Ответ: $\\angle BCO = 32°$
` + }, + { + text: `В бензобак грузового автомобиля МАЗ залили $200$ л бензина. + В каждый день пути расходовалось $m$ литров бензина. + На сколько дней хватит $200$ л бензина? + Составьте формулу зависимости числа дней $k$ от количества литров бензина, + расходуемого каждый день.`, + sol: `Метод составления уравнения по условию задачи. +
Шаг 1. Введём обозначение. Пусть $k$ — искомое число дней, на которые хватит бензина. +
Шаг 2. По условию за один день расходуется $m$ литров. Значит, за $k$ дней израсходуется $m\\cdot k$ литров. +
Шаг 3. Так как в баке всего $200$ л и весь бензин будет израсходован за $k$ дней, составим уравнение: +$$m\\cdot k = 200$$ +Шаг 4. Выразим $k$. Разделим обе части на $m$ (это можно сделать, так как $m\\ne 0$): +$$k = \\dfrac{200}{m}$$ +
Ответ: бензина хватит на $\\dfrac{200}{m}$ дней;  формула: $k = \\dfrac{200}{m}$
` + }, + { + text: `Решите уравнение $\\dfrac{3}{8} : \\dfrac{3}{4} = y : \\dfrac{2}{3}$.`, + sol: `Правило деления дробей: чтобы разделить на дробь, надо умножить на дробь, обратную делителю: $\\dfrac{a}{b} : \\dfrac{c}{d} = \\dfrac{a}{b} \\cdot \\dfrac{d}{c}$. +
Шаг 1. Вычислим левую часть уравнения: +$$\\dfrac{3}{8}:\\dfrac{3}{4} = \\dfrac{3}{8}\\cdot\\dfrac{4}{3} = \\dfrac{12}{24} = \\dfrac{1}{2}$$ +Шаг 2. Уравнение принимает вид: +$$\\dfrac{1}{2} = y : \\dfrac{2}{3}$$ +По правилу деления, $y : \\dfrac{2}{3} = y \\cdot \\dfrac{3}{2}$. Значит: +$$\\dfrac{1}{2} = y \\cdot \\dfrac{3}{2}$$ +Шаг 3. Находим $y$. Умножим обе части на $\\dfrac{2}{3}$: +$$y = \\dfrac{1}{2} \\cdot \\dfrac{2}{3} = \\dfrac{1}{3}$$ +Проверка: $\\dfrac{1}{3} : \\dfrac{2}{3} = \\dfrac{1}{3} \\cdot \\dfrac{3}{2} = \\dfrac{1}{2}$ ✓ +
Ответ: $y = \\dfrac{1}{3}$
` + }, + { + text: `Из Фаниполя в Юцковские родники, расстояние между которыми равно $20$ км, вышел турист. + Одновременно с ним по тому же маршруту из Юцковских родников в Фаниполь выехал велосипедист, + скорость которого в $4$ раза больше скорости пешехода. + Сколько километров осталось преодолеть туристу до Юцковских родников + после встречи с велосипедистом?`, + sol: `Метод введения переменной и составления уравнения движения навстречу. При движении навстречу скорости участников складываются: $v_{\\text{сбл}} = v_1 + v_2$. +
Шаг 1. Пусть скорость пешехода равна $v$ км/ч. По условию скорость велосипедиста в $4$ раза больше, значит, она равна $4v$ км/ч. +
Шаг 2. Так как они движутся навстречу друг другу, скорость сближения равна сумме скоростей: +$$v_{\\text{сбл}} = v + 4v = 5v\\text{ км/ч}$$ +
Шаг 3. Пусть $t$ — время от старта до встречи. За это время участники вместе прошли всё расстояние $20$ км: +$$5v\\cdot t = 20 \\implies v\\cdot t = 4$$ +Значит, путь $v\\cdot t$, пройденный туристом до встречи, равен $4$ км. +
Шаг 4. Так как турист шёл из Фаниполя, ему осталось пройти: +$$20 - 4 = 16\\text{ км}$$ +
Ответ: $16$ км
` + }, + { + text: `Основания трапеции равны $10$ см и $35$ см, боковые стороны — $15$ см и $20$ см. + Найдите площадь трапеции.`, + sol: `Формула площади трапеции: $S = \\dfrac{a+b}{2}\\cdot h$, где $a$ и $b$ — основания, $h$ — высота. +
Чтобы найти высоту, опустим из вершин верхнего основания перпендикуляры $BH$ и $CK$ на нижнее основание. + + + + + + + A + B + C + D + H + K + 10 + 35 + 15 + 20 + h + 9 + 16 + 10 + +Шаг 1. Так как $BH\\parallel CK$ и оба перпендикулярны $AD$, четырёхугольник $BCKH$ — прямоугольник. Значит, $HK = BC = 10$ см. +
Тогда сумма «выступов» по краям: +$$AH + KD = AD - HK = 35 - 10 = 25\\text{ см}$$ +Шаг 2. По теореме Пифагора ($c^2 = a^2 + b^2$) в прямоугольных треугольниках $ABH$ и $DCK$: +$$AH^2 + h^2 = AB^2 = 15^2 = 225$$ +$$KD^2 + h^2 = CD^2 = 20^2 = 400$$ +Шаг 3. Вычтем первое равенство из второго: +$$KD^2 - AH^2 = 400 - 225 = 175$$ +По формуле разности квадратов ($a^2 - b^2 = (a-b)(a+b)$): +$$(KD+AH)(KD-AH) = 175$$ +Поскольку $KD + AH = 25$: +$$25\\cdot(KD-AH) = 175 \\implies KD - AH = 7$$ +Шаг 4. Из системы $\\{AH + KD = 25;\\; KD - AH = 7\\}$ получаем: $AH = 9$, $KD = 16$. +
Шаг 5. Находим высоту $h$ по теореме Пифагора: +$$h^2 = 225 - AH^2 = 225 - 81 = 144 \\implies h = 12\\text{ см}$$ +Шаг 6. Подставляем в формулу площади: +$$S = \\dfrac{10+35}{2}\\cdot 12 = \\dfrac{45}{2}\\cdot 12 = 270\\text{ см}^2$$ +
Ответ: $270$ см²
` + }, + { + text: `Известно, что $x_1$ и $x_2$ — корни уравнения $0{,}1x^2 + 0{,}7x - 1{,}2 = 0$. + Найдите значение выражения $\\dfrac{x_1 x_2}{-3x_1^2 - 3x_2^2}$.`, + sol: `Теорема Виета: для приведённого уравнения $x^2 + px + q = 0$ корни $x_1$ и $x_2$ удовлетворяют: +$$x_1 + x_2 = -p, \\quad x_1 \\cdot x_2 = q$$ +Шаг 1. Приведём уравнение к виду $x^2 + px + q = 0$. Разделим обе части на $0{,}1$: +$$x^2 + 7x - 12 = 0$$ +Шаг 2. По теореме Виета: +$$x_1 + x_2 = -7, \\quad x_1 \\cdot x_2 = -12$$ +Шаг 3. Чтобы найти $x_1^2 + x_2^2$, воспользуемся формулой квадрата суммы: +$$(x_1 + x_2)^2 = x_1^2 + 2x_1 x_2 + x_2^2$$ +Отсюда: +$$x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = (-7)^2 - 2\\cdot(-12) = 49 + 24 = 73$$ +Шаг 4. Подставляем в исходное выражение: +$$\\dfrac{x_1 x_2}{-3x_1^2 - 3x_2^2} = \\dfrac{x_1 x_2}{-3(x_1^2 + x_2^2)} = \\dfrac{-12}{-3\\cdot 73} = \\dfrac{12}{219} = \\dfrac{4}{73}$$ +
Ответ: $\\dfrac{4}{73}$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v34.js b/frontend/js/exam9/variants/v34.js new file mode 100644 index 0000000..eeaf6f0 --- /dev/null +++ b/frontend/js/exam9/variants/v34.js @@ -0,0 +1,186 @@ +VARIANTS[34] = { + label: "Вариант 34", + tasks: [ + { + text: `Отношение чисел $24$ и $4$ равно:`, + opts: [ + ["а", "$18$"], ["б", "$32$"], ["в", "$6$"], + ["г", "$14$"], ["д", "$1$"], + ], + sol: `$$24 : 4 = 6$$ +
Ответ: в) $6$
` + }, + { + text: `Запись выражения $1 - \\dfrac{a}{b}$ в виде дроби имеет вид:`, + opts: [ + ["а", "$\\dfrac{1-a}{b}$"], ["б", "$\\dfrac{b-a}{b}$"], ["в", "$\\dfrac{a-b}{b}$"], + ["г", "$b - a$"], ["д", "$\\dfrac{a-1}{b}$"], + ], + sol: `Приводим к общему знаменателю $b$: +$$1 - \\dfrac{a}{b} = \\dfrac{b}{b} - \\dfrac{a}{b} = \\dfrac{b-a}{b}$$ +
Ответ: б) $\\dfrac{b-a}{b}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "если четырёхугольник $ABCD$ описан около окружности, то $BC + AD = AB + CD$;"], + ["б", "котангенс острого угла прямоугольного треугольника равен отношению прилежащего катета к противолежащему;"], + ["в", "вписанный угол равен половине соответствующего центрального угла;"], + ["г", "любая биссектриса равнобедренного треугольника является его медианой?"], + ], + sol: `
    +
  • а) Для описанного четырёхугольника: $BC+AD=AB+CD$ — верно (свойство описанного четырёхугольника)
    • +
  • б) $\\ctg\\alpha = \\dfrac{\\text{прилежащий катет}}{\\text{противолежащий катет}}$ — верно
    • +
  • в) Вписанный угол равен половине соответствующего центрального угла — верно
    • +
  • г) «Любая биссектриса равнобедренного треугольника является его медианой» — НЕВЕРНО. Биссектриса из вершинного угла к основанию является медианой, а биссектрисы боковых углов медианами не являются.
    • +
+
Ответ: г)
` + }, + { + text: `Решите неравенство $\\dfrac{x}{2} + 4 < 0$ и укажите наибольшее целое решение этого неравенства.`, + sol: `$$\\dfrac{x}{2} \\lt -4 \\implies x \\lt -8$$ +Решение: $x\\in(-\\infty;\\,-8)$. Наибольшее целое число, строго меньшее $-8$ — это $\\mathbf{-9}$. +
Ответ: $x \\lt -8$; наибольшее целое решение $= -9$
` + }, + { + text: `В угол $B$ вписана окружность с центром в точке $O$, которая касается сторон угла + в точках $A$ и $C$. Найдите угол $ABO$, если $\\angle AOC = 118^{\\circ}$.`, + sol: `Свойство касательной к окружности: радиус, проведённый в точку касания, перпендикулярен касательной. +
Значит, $\\angle OAB = 90°$ и $\\angle OCB = 90°$. + + + + + + + + + + + + + + B + O + C + A + 62° + 118° + +Шаг 1. Найдём угол $B$ из четырёхугольника $ABOC$. +
По теореме о сумме углов четырёхугольника: сумма углов любого четырёхугольника равна $360°$. +$$\\angle B + \\angle OAB + \\angle AOC + \\angle OCB = 360°$$ +Подставляем известные углы: +$$\\angle B + 90° + 118° + 90° = 360°$$ +$$\\angle B = 360° - 298° = 62°$$ +Шаг 2. Применим свойство биссектрисы: центр окружности, вписанной в угол, лежит на биссектрисе этого угла (так как $OA = OC$ — оба радиуса). +
Значит, $BO$ — биссектриса угла $B$, и она делит его пополам: +$$\\angle ABO = \\dfrac{\\angle B}{2} = \\dfrac{62°}{2} = 31°$$ +
Ответ: $\\angle ABO = 31°$
` + }, + { + text: `Путешественники залили в бензобак автомобиля Geely Atlas $58$ л бензина. + В каждый день пути расходовалось $p$ литров бензина. + На сколько дней путешественникам хватит бензина? + Составьте формулу зависимости количества дней $k$ от количества литров бензина, + расходуемого каждый день.`, + sol: `Метод составления уравнения по условию задачи. +
Шаг 1. Введём обозначение. Пусть $k$ — искомое число дней, на которые хватит бензина. +
Шаг 2. По условию за один день расходуется $p$ литров. Значит, за $k$ дней израсходуется $p\\cdot k$ литров. +
Шаг 3. Так как в баке всего $58$ л и весь бензин будет израсходован за $k$ дней, составим уравнение: +$$p\\cdot k = 58$$ +Шаг 4. Выразим $k$. Разделим обе части на $p$ (это можно сделать, так как $p\\ne 0$): +$$k = \\dfrac{58}{p}$$ +
Ответ: бензина хватит на $\\dfrac{58}{p}$ дней;  формула: $k = \\dfrac{58}{p}$
` + }, + { + text: `Решите уравнение $\\dfrac{4x+2}{3} = \\dfrac{1}{3}$.`, + sol: `Свойство пропорции (или равенства дробей с одинаковыми знаменателями): если $\\dfrac{a}{c} = \\dfrac{b}{c}$, то $a = b$. +
Шаг 1. Знаменатели в обеих частях одинаковые ($=3$), поэтому числители равны: +$$4x + 2 = 1$$ +Шаг 2. Переносим $2$ в правую часть (с противоположным знаком): +$$4x = 1 - 2 = -1$$ +Шаг 3. Делим обе части на $4$: +$$x = -\\dfrac{1}{4}$$ +Проверка: $\\dfrac{4\\cdot\\left(-\\dfrac{1}{4}\\right)+2}{3} = \\dfrac{-1+2}{3} = \\dfrac{1}{3}$ ✓ +
Ответ: $x = -\\dfrac{1}{4}$
` + }, + { + text: `Из Жодино в Радошковичи, расстояние между которыми равно $60$ км, выехал мотоциклист. + Одновременно с ним по тому же маршруту из Радошковичей в Жодино выехал велосипедист, + скорость которого в $5$ раз меньше скорости мотоциклиста. + Сколько километров осталось преодолеть мотоциклисту до Радошковичей + после встречи с велосипедистом?`, + sol: `Метод введения переменной и составления уравнения движения навстречу. При движении навстречу скорости участников складываются: $v_{\\text{сбл}} = v_1 + v_2$. +
Шаг 1. Пусть скорость мотоциклиста равна $v$ км/ч. По условию скорость велосипедиста в $5$ раз меньше, значит, она равна $\\dfrac{v}{5}$ км/ч. +
Шаг 2. Так как они движутся навстречу друг другу, скорость сближения равна сумме скоростей: +$$v_{\\text{сбл}} = v + \\dfrac{v}{5} = \\dfrac{6v}{5}\\text{ км/ч}$$ +
Шаг 3. Пусть $t$ — время от старта до встречи. За это время участники вместе преодолели всё расстояние $60$ км: +$$\\dfrac{6v}{5}\\cdot t = 60 \\implies v\\cdot t = 50$$ +Значит, путь $v\\cdot t$, проделанный мотоциклистом до встречи, равен $50$ км. +
Шаг 4. Так как мотоциклист ехал из Жодино, ему осталось: +$$60 - 50 = 10\\text{ км}$$ +
Ответ: $10$ км
` + }, + { + text: `Основания трапеции равны $5$ см и $15$ см, боковые стороны — $6$ см и $8$ см. + Найдите площадь трапеции.`, + sol: `Формула площади трапеции: $S = \\dfrac{a+b}{2}\\cdot h$, где $a$, $b$ — основания, $h$ — высота. +
Чтобы найти высоту, опустим из вершин $B$ и $C$ верхнего основания перпендикуляры $BH$ и $CK$ на нижнее основание. + + + + + + + A + B + C + D + H + K + 5 + 15 + 6 + 8 + h + 3,6 + 6,4 + 5 + +Шаг 1. Так как $BH\\parallel CK$ и оба перпендикулярны $AD$, четырёхугольник $BCKH$ — прямоугольник. Значит, $HK = BC = 5$ см. +
Тогда сумма «выступов» по краям: +$$AH + KD = AD - HK = 15 - 5 = 10\\text{ см}$$ +Шаг 2. По теореме Пифагора ($c^2 = a^2 + b^2$) в прямоугольных треугольниках $ABH$ и $DCK$: +$$AH^2 + h^2 = AB^2 = 6^2 = 36$$ +$$KD^2 + h^2 = CD^2 = 8^2 = 64$$ +Шаг 3. Вычтем первое равенство из второго: +$$KD^2 - AH^2 = 64 - 36 = 28$$ +По формуле разности квадратов: +$$(KD+AH)(KD-AH) = 28$$ +Поскольку $KD + AH = 10$: +$$10\\cdot(KD - AH) = 28 \\implies KD - AH = 2{,}8$$ +Шаг 4. Из системы $\\{AH + KD = 10;\\; KD - AH = 2{,}8\\}$ получаем: $AH = 3{,}6$, $KD = 6{,}4$. +
Шаг 5. Находим высоту $h$ по теореме Пифагора: +$$h^2 = 36 - 3{,}6^2 = 36 - 12{,}96 = 23{,}04 \\implies h = 4{,}8\\text{ см}$$ +Шаг 6. Подставляем в формулу площади: +$$S = \\dfrac{5+15}{2}\\cdot 4{,}8 = 10\\cdot 4{,}8 = 48\\text{ см}^2$$ +
Ответ: $48$ см²
` + }, + { + text: `Известно, что $x_1$ и $x_2$ — корни уравнения $x^2 - 3x - 5 = 0$. + Найдите значение выражения $\\dfrac{2x_1 x_2}{-5x_1^2 - 5x_2^2}$.`, + sol: `Теорема Виета: для приведённого уравнения $x^2 + px + q = 0$ корни $x_1$ и $x_2$ удовлетворяют: +$$x_1 + x_2 = -p, \\quad x_1 \\cdot x_2 = q$$ +Шаг 1. Применяем теорему Виета к уравнению $x^2 - 3x - 5 = 0$ (здесь $p = -3$, $q = -5$): +$$x_1 + x_2 = 3, \\quad x_1 \\cdot x_2 = -5$$ +Шаг 2. Чтобы найти $x_1^2 + x_2^2$, воспользуемся формулой квадрата суммы: +$$(x_1 + x_2)^2 = x_1^2 + 2x_1 x_2 + x_2^2$$ +Отсюда: +$$x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = 3^2 - 2\\cdot(-5) = 9 + 10 = 19$$ +Шаг 3. Подставляем в исходное выражение: +$$\\dfrac{2x_1 x_2}{-5x_1^2 - 5x_2^2} = \\dfrac{2x_1 x_2}{-5(x_1^2 + x_2^2)} = \\dfrac{2\\cdot(-5)}{-5\\cdot 19} = \\dfrac{-10}{-95} = \\dfrac{2}{19}$$ +
Ответ: $\\dfrac{2}{19}$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v35.js b/frontend/js/exam9/variants/v35.js new file mode 100644 index 0000000..0079eb7 --- /dev/null +++ b/frontend/js/exam9/variants/v35.js @@ -0,0 +1,254 @@ +VARIANTS[35] = { + label: "Вариант 35", + tasks: [ + { + text: `При каком значении переменной выражение $\\dfrac{m}{n-2}$ НЕ имеет смысла:`, + opts: [ + ["а", "$-2$"], ["б", "$-1$"], ["в", "$0$"], ["г", "$1$"], ["д", "$2$"], + ], + sol: `Дробь не имеет смысла при нулевом знаменателе: +$$n - 2 = 0 \\implies n = 2$$ +
Ответ: д) $2$
` + }, + { + text: `Если к разности чисел $-1{,}2$ и $0{,}6$ прибавить $1{,}8$, то получится число:`, + opts: [ + ["а", "$-0{,}9$"], ["б", "$-0{,}8$"], ["в", "$0$"], ["г", "$1{,}2$"], ["д", "$3{,}6$"], + ], + sol: `$$(-1{,}2 - 0{,}6) + 1{,}8 = -1{,}8 + 1{,}8 = 0$$ +
Ответ: в) $0$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "около четырёхугольника $ABCD$, где $\\angle A = 40^{\\circ}$, $\\angle C = 140^{\\circ}$, можно описать окружность;"], + ["б", "$\\sin 30^{\\circ} = \\sin 150^{\\circ}$;"], + ["в", "вписанный угол, опирающийся на диаметр, — прямой;"], + ["г", "в любом равнобедренном треугольнике все высоты равны между собой?"], + ], + sol: `
    +
  • а) $\\angle A + \\angle C = 40°+140°=180°$ ⟹ около него можно описать окружность — верно
    • +
  • б) $\\sin30°=\\sin(180°-30°)=\\sin150°=0{,}5$ — верно
    • +
  • в) Вписанный угол на диаметре $= 90°$ — верно
    • +
  • г) «Все высоты равнобедренного треугольника равны» — НЕВЕРНО. В равнобедренном (но не равностороннем) треугольнике две высоты к боковым сторонам равны, но высота к основанию — другая.
    • +
+
Ответ: г)
` + }, + { + text: `Из $4$ кг муки получается $3{,}2$ кг печенья. + Сколько надо муки, чтобы испечь $2{,}4$ кг печенья?`, + sol: `Метод прямой пропорции: если две величины прямо пропорциональны, то отношение их соответствующих значений равно: $\\dfrac{a_1}{b_1} = \\dfrac{a_2}{b_2}$. +
Шаг 1. Обозначим $x$ — искомое количество муки (в кг), необходимое для $2{,}4$ кг печенья. +
Шаг 2. Так как количество печенья прямо пропорционально количеству муки (чем больше муки — тем больше печенья), составим пропорцию: +$$\\dfrac{4\\text{ кг муки}}{3{,}2\\text{ кг печенья}} = \\dfrac{x}{2{,}4\\text{ кг печенья}}$$ +Шаг 3. По основному свойству пропорции произведение крайних членов равно произведению средних, откуда: +$$x = \\dfrac{4\\cdot 2{,}4}{3{,}2} = \\dfrac{9{,}6}{3{,}2} = 3\\text{ кг}$$ +
Ответ: $3$ кг муки
` + }, + { + text: `Найдите $\\text{НОК}(14;\\; 42;\\; 336)$. + В ответ запишите число, обратное полученному.`, + sol: `Разложим на простые множители: +$$14 = 2\\cdot7, \\quad 42 = 2\\cdot3\\cdot7, \\quad 336 = 2^4\\cdot3\\cdot7$$ +НОК берёт наибольшие степени каждого простого: +$$\\text{НОК} = 2^4\\cdot3\\cdot7 = 16\\cdot21 = 336$$ +Обратное число: $\\dfrac{1}{336}$. +
Ответ: $\\dfrac{1}{336}$
` + }, + { + text: `$ABCD$ — параллелограмм, $BK$ — высота, проведённая к стороне $AD$, + $CD = 10$ см, $KD = 7$ см, $\\angle A = 60^{\\circ}$. + Найдите периметр параллелограмма.`, + sol: `Свойство параллелограмма: противоположные стороны параллелограмма равны. +
Значит, $AB = CD = 10$ см. + + + + + + + A + B + C + D + K + 60° + 10 + AK=5 + KD=7 + h + +Шаг 1. Рассмотрим прямоугольный треугольник $ABK$ ($\\angle BKA = 90°$, $\\angle A = 60°$, $AB = 10$ см). +
По определению косинуса острого угла прямоугольного треугольника: $\\cos\\alpha = \\dfrac{\\text{прилежащий катет}}{\\text{гипотенуза}}$. +
Здесь прилежащий к углу $A$ катет — $AK$, гипотенуза — $AB$: +$$\\cos 60° = \\dfrac{AK}{AB} \\implies AK = AB\\cdot\\cos 60° = 10\\cdot\\dfrac{1}{2} = 5\\text{ см}$$ +Шаг 2. Находим длину стороны $AD$: +$$AD = AK + KD = 5 + 7 = 12\\text{ см}$$ +Шаг 3. Формула периметра параллелограмма: $P = 2(a + b)$, где $a$, $b$ — соседние стороны. +$$P = 2(AB + AD) = 2(10 + 12) = 44\\text{ см}$$ +
Ответ: $44$ см
` + }, + { + text: `Сократите дробь $\\dfrac{b^2 + 15b + 56}{b^2 + 3b - 28}$ + и найдите значение полученного выражения при $b = -6$.`, + sol: `Теорема Виета (для разложения квадратного трёхчлена): $x^2 + px + q = (x - x_1)(x - x_2)$, где $x_1$, $x_2$ — корни уравнения $x^2 + px + q = 0$, причём $x_1 + x_2 = -p$, $x_1\\cdot x_2 = q$. +
Шаг 1. Разложим числитель $b^2 + 15b + 56$. +
Подберём такие два числа, сумма которых $= -15$ (точнее, $= 15$ для $-p$), а произведение $= 56$. Это $7$ и $8$ (так как $7 + 8 = 15$ и $7\\cdot 8 = 56$): +$$b^2 + 15b + 56 = (b + 7)(b + 8)$$ +Шаг 2. Разложим знаменатель $b^2 + 3b - 28$. +
Подберём числа с суммой $-3$ и произведением $-28$. Это $7$ и $-4$ (так как $7 + (-4) = 3$, а $7\\cdot(-4) = -28$): +$$b^2 + 3b - 28 = (b + 7)(b - 4)$$ +Шаг 3. Сокращаем общий множитель $(b + 7)$, считая $b \\neq -7$ и $b \\neq 4$: +$$\\dfrac{(b+7)(b+8)}{(b+7)(b-4)} = \\dfrac{b+8}{b-4}$$ +Шаг 4. Подставляем $b = -6$ в полученное выражение: +$$\\dfrac{-6+8}{-6-4} = \\dfrac{2}{-10} = -\\dfrac{1}{5}$$ +
Ответ: $\\dfrac{b+8}{b-4}$;  при $b=-6$ значение равно $-\\dfrac{1}{5}$
` + }, + { + text: `Найдите сумму всех натуральных чисел, больших $12$ и не превосходящих $121$, + которые при делении на $6$ дают в остатке $1$.`, + sol: `Формула суммы $n$ членов арифметической прогрессии: $S_n = \\dfrac{n(a_1 + a_n)}{2}$, где $a_1$ — первый член, $a_n$ — последний, $n$ — число членов. +
Шаг 1. По определению деления с остатком число вида «$6k+1$» при делении на $6$ даёт остаток $1$. Значит, нужные числа имеют вид $a = 6k+1$, где $k$ — целое неотрицательное. +
Шаг 2. Найдём допустимые значения $k$ из условия $12 \\lt a \\leq 121$: +$$12 \\lt 6k+1 \\leq 121 \\implies 11 \\lt 6k \\leq 120 \\implies \\dfrac{11}{6} \\lt k \\leq 20$$ +Так как $k$ целое, получаем $k = 2,\\ 3,\\ \\ldots,\\ 20$. +
Шаг 3. Найдём первый и последний члены последовательности: +$$a_1 = 6\\cdot 2 + 1 = 13, \\quad a_n = 6\\cdot 20 + 1 = 121$$ +Шаг 4. Числа $13,\\ 19,\\ 25,\\ \\ldots,\\ 121$ образуют арифметическую прогрессию с разностью $d = 6$ (каждое следующее больше предыдущего на $6$). +
Шаг 5. Найдём количество членов по формуле $n = \\dfrac{a_n - a_1}{d} + 1$: +$$n = \\dfrac{121 - 13}{6} + 1 = 18 + 1 = 19$$ +Шаг 6. Применяем формулу суммы: +$$S = \\dfrac{n(a_1 + a_n)}{2} = \\dfrac{19\\cdot(13 + 121)}{2} = \\dfrac{19\\cdot 134}{2} = 19\\cdot 67 = 1273$$ +
Ответ: $1273$
` + }, + { + text: `Квадратный участок земли разбили на четыре части: газон, цветник, огород и сад. + Сад и цветник — квадраты. Периметр сада — $84$ м, а цветника — $24$ м. + Чему равен периметр газона?`, + sol: `Формула периметра квадрата: $P = 4a$, откуда сторона $a = \\dfrac{P}{4}$. +
Шаг 1. Находим стороны квадратных частей: +
— Сторона сада: $84 \\div 4 = 21$ м. +
— Сторона цветника: $24 \\div 4 = 6$ м. +
Шаг 2. Участок разбит одной горизонтальной и одной вертикальной линией на 4 прямоугольные части. Сад и цветник — квадраты, значит их стороны определяют, как поделён участок: + + + + + + + + + Газон + Цветник + Сад + Огород + 21 м + 6 м + 21 м + 21 м + 6 м + +Шаг 3. Сторона всего квадратного участка $= 21 + 6 = 27$ м. +
Шаг 4. Газон — прямоугольник со сторонами $21$ м и $6$ м (по рисунку). +
По формуле периметра прямоугольника $P = 2(a + b)$: +$$P_{\\text{газон}} = 2(21 + 6) = 2\\cdot 27 = 54\\text{ м}$$ +
Ответ: $54$ м
` + }, + { + text: `Дан треугольник $ABC$ со сторонами $AC = 6$, $BC = 8$, $AB = 10$. + Найдите расстояние между центрами описанной и вписанной окружностей треугольника.`, + sol: ` + + + + + + + + + + + + + + + + + + + + + + + + + + C + A + B + + O + I + + 3 + 4 + + 2 + 2 + + 1 + 2 + + √5 + + 6 + 8 + 10 + + R = 5 + r = 2 + +Шаг 1. Тип треугольника. +$$6^2+8^2 = 36+64 = 100 = 10^2 \\checkmark$$ +Треугольник прямоугольный — прямой угол при $C$. +
Шаг 2. Центр и радиус описанной окружности $O$. +
В прямоугольном треугольнике гипотенуза — диаметр описанной окружности. +Значит, центр $O$ — это просто середина гипотенузы $AB$: +$$R = \\dfrac{AB}{2} = \\dfrac{10}{2} = 5\\text{ см}$$ +Шаг 3. Центр и радиус вписанной окружности $I$. +
Вписанная окружность касается всех трёх сторон. Её радиус: +$$r = \\dfrac{AC + BC - AB}{2} = \\dfrac{6+8-10}{2} = 2\\text{ см}$$ +Центр $I$ стоит на расстоянии $r=2$ от каждой стороны треугольника. +
Шаг 4. Как далеко $O$ и $I$ от катетов? +
Смотрим на рисунок — пунктирные линии от $O$ и $I$ до катетов. +

+Центр $O$ — середина $AB$. Смотрим, как далеко вершины от катета $BC$: +
    +
  • Вершина $A$ — на расстоянии $AC = 6$ от $BC$.
    • +
  • Вершина $B$ — прямо на $BC$, расстояние $= 0$.
    • +
  • Середина $O$ — посередине: $(6 + 0) : 2 =$ 3 от $BC$.
    • +
+Теперь смотрим, как далеко от катета $AC$: +
    +
  • Вершина $A$ — прямо на $AC$, расстояние $= 0$.
    • +
  • Вершина $B$ — на расстоянии $BC = 8$ от $AC$.
    • +
  • Середина $O$ — посередине: $(0 + 8) : 2 =$ 4 от $AC$.
    • +
+Центр $I$ — по определению на расстоянии $r=2$ от каждой стороны: +расстояние до $BC$ = 2, до $AC$ = 2. +

+Итог таблицей: + + + + + +
до катета BCдо катета AC
O34
I22
разность3−2 = 14−2 = 2
+Шаг 5. Находим $OI$. +
Разности — это катеты прямоугольного треугольника (красный на рисунке) между $O$ и $I$. +
По теореме Пифагора: +$$OI = \\sqrt{1^2 + 2^2} = \\sqrt{1+4} = \\sqrt{5}\\text{ см}$$ +
Ответ: $\\sqrt{5}$ см
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v36.js b/frontend/js/exam9/variants/v36.js new file mode 100644 index 0000000..95c5d62 --- /dev/null +++ b/frontend/js/exam9/variants/v36.js @@ -0,0 +1,211 @@ +VARIANTS[36] = { + label: "Вариант 36", + tasks: [ + { + text: `При каком значении переменной выражение $\\dfrac{a}{a+3}$ НЕ имеет смысла:`, + opts: [ + ["а", "$-3$"], ["б", "$-2$"], ["в", "$0$"], ["г", "$2$"], ["д", "$3$"], + ], + sol: `Дробь не имеет смысла при нулевом знаменателе: +$$a + 3 = 0 \\implies a = -3$$ +
Ответ: а) $-3$
` + }, + { + text: `Если к разности чисел $1{,}2$ и $-0{,}6$ прибавить $0{,}8$, то получится число:`, + opts: [ + ["а", "$-1{,}4$"], ["б", "$1$"], ["в", "$-1{,}2$"], ["г", "$2{,}6$"], ["д", "$1{,}6$"], + ], + sol: `$$(1{,}2 - (-0{,}6)) + 0{,}8 = (1{,}2 + 0{,}6) + 0{,}8 = 1{,}8 + 0{,}8 = 2{,}6$$ +
Ответ: г) $2{,}6$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "около четырёхугольника $ABCD$, где $\\angle B = 30^{\\circ}$, $\\angle D = 150^{\\circ}$, можно описать окружность;"], + ["б", "$\\cos 60^{\\circ} = -\\cos 120^{\\circ}$;"], + ["в", "прямой вписанный угол опирается на диаметр;"], + ["г", "в любом равнобедренном треугольнике все медианы равны между собой?"], + ], + sol: `
    +
  • а) $\\angle B + \\angle D = 30°+150°=180°$ ⟹ около него можно описать окружность — верно
    • +
  • б) $\\cos60°=0{,}5$; $-\\cos120°=-(-0{,}5)=0{,}5$ ⟹ равны — верно
    • +
  • в) Прямой вписанный угол ($90°$) опирается на диаметр — верно
    • +
  • г) «Все медианы равнобедренного треугольника равны» — НЕВЕРНО. В равнобедренном (но не равностороннем) треугольнике две медианы к боковым сторонам равны, а медиана к основанию — другая.
    • +
+
Ответ: г)
` + }, + { + text: `Из $4$ кг свежих яблок получается $0{,}5$ кг сушёных. + Сколько килограммов свежих яблок надо взять, чтобы получить $4$ кг сушёных?`, + sol: `Метод прямой пропорции: если две величины прямо пропорциональны, то $\\dfrac{a_1}{b_1} = \\dfrac{a_2}{b_2}$. +
Шаг 1. Обозначим $x$ — искомое количество свежих яблок (в кг), необходимое для получения $4$ кг сушёных. +
Шаг 2. Так как масса сушёных яблок прямо пропорциональна массе свежих (чем больше свежих, тем больше получится сушёных), составим пропорцию: +$$\\dfrac{4\\text{ кг свежих}}{0{,}5\\text{ кг сушёных}} = \\dfrac{x}{4\\text{ кг сушёных}}$$ +Шаг 3. По основному свойству пропорции (произведение крайних членов равно произведению средних): +$$x = \\dfrac{4\\cdot 4}{0{,}5} = \\dfrac{16}{0{,}5} = 32\\text{ кг}$$ +
Ответ: $32$ кг свежих яблок
` + }, + { + text: `Найдите $\\text{НОК}(21;\\; 63;\\; 105)$. + В ответ запишите число, обратное полученному.`, + sol: `Разложим на простые множители: +$$21 = 3\\cdot7, \\quad 63 = 3^2\\cdot7, \\quad 105 = 3\\cdot5\\cdot7$$ +НОК берёт наибольшие степени каждого простого: +$$\\text{НОК} = 3^2\\cdot5\\cdot7 = 9\\cdot35 = 315$$ +Обратное число: $\\dfrac{1}{315}$. +
Ответ: $\\dfrac{1}{315}$
` + }, + { + text: `$ABCD$ — параллелограмм, $CK$ — высота, проведённая к стороне $AD$, + $BC = 12$ см, $AK = 8$ см, $\\angle A = 120^{\\circ}$. + Найдите периметр параллелограмма.`, + sol: ` + + + + + + A + B + C + D + K + 120° + 8 + 12 + AK=8 + h + +Свойство параллелограмма: противоположные стороны параллелограмма равны. +
Значит, $AD = BC = 12$ см. +
Шаг 1. Так как угол $A$ тупой ($120°$), основание высоты $K$, опущенной из вершины $C$ на прямую $AD$, лежит за точкой $D$, то есть $AK \\gt AD$ (см. рисунок: $AK = 8$, $AD = 12$ означает, что $K$ ближе к $A$, чем $D$, на $4$ см). +
Точнее: смежный угол при вершине $D$ равен $180° - 120° = 60°$. +
Шаг 2. Рассмотрим прямоугольный треугольник $CKD$ ($\\angle CKD = 90°$). +
Угол $CDK = 180° - \\angle ADC = 180° - 60° = 120°$? Нет — высота $CK$ из $C$ падает на сторону $AD$ изнутри параллелограмма. Угол $\\angle CDA$ у параллелограмма (противоположен $\\angle B$, равен $60°$, так как соседние углы дают $180°$). +
В $\\triangle CKD$: $\\cos\\angle CDK = \\dfrac{KD}{CD}$, при этом $KD = AD - AK = 12 - 8 = 4$ см. +
По определению косинуса: +$$\\cos 60° = \\dfrac{KD}{CD} \\implies \\dfrac{1}{2} = \\dfrac{4}{CD} \\implies CD = 8\\text{ см}$$ +Шаг 3. $AB = CD = 8$ см (противоположные стороны). +
Шаг 4. Формула периметра параллелограмма: $P = 2(a + b)$: +$$P = 2(AB + BC) = 2(8 + 12) = 40\\text{ см}$$ +
Ответ: $40$ см
` + }, + { + text: `Сократите дробь $\\dfrac{t^2 + 9t - 22}{t^2 + 7t - 44}$ + и найдите значение полученного выражения при $t = -6$.`, + sol: `Теорема Виета (для разложения квадратного трёхчлена): $x^2 + px + q = (x - x_1)(x - x_2)$, где $x_1$, $x_2$ — корни уравнения $x^2 + px + q = 0$. +
Шаг 1. Разложим числитель $t^2 + 9t - 22$. +
Ищем два числа с суммой $-9$ (по Виете $x_1 + x_2 = -p = -9$) и произведением $-22$. Это $-11$ и $2$ (так как $-11 + 2 = -9$, $-11 \\cdot 2 = -22$): +$$t^2 + 9t - 22 = (t + 11)(t - 2)$$ +Шаг 2. Разложим знаменатель $t^2 + 7t - 44$. +
Ищем два числа с суммой $-7$ и произведением $-44$. Это $-11$ и $4$ (так как $-11 + 4 = -7$, $-11 \\cdot 4 = -44$): +$$t^2 + 7t - 44 = (t + 11)(t - 4)$$ +Шаг 3. Сокращаем общий множитель $(t + 11)$, считая $t \\neq -11$ и $t \\neq 4$: +$$\\dfrac{(t+11)(t-2)}{(t+11)(t-4)} = \\dfrac{t-2}{t-4}$$ +Шаг 4. Подставляем $t = -6$ в полученное выражение: +$$\\dfrac{-6-2}{-6-4} = \\dfrac{-8}{-10} = \\dfrac{4}{5}$$ +
Ответ: $\\dfrac{t-2}{t-4}$;  при $t=-6$ значение равно $\\dfrac{4}{5}$
` + }, + { + text: `Найдите сумму всех натуральных чисел, больших $8$ и не превосходящих $188$, + которые при делении на $8$ дают в остатке $4$.`, + sol: `Формула суммы $n$ членов арифметической прогрессии: $S_n = \\dfrac{n(a_1 + a_n)}{2}$. +
Шаг 1. По определению деления с остатком натуральные числа, дающие при делении на $8$ остаток $4$, имеют вид $a = 8k + 4$, где $k$ — целое неотрицательное. +
Шаг 2. Найдём допустимые значения $k$ из условия $8 \\lt a \\leq 188$: +$$8 \\lt 8k + 4 \\leq 188 \\implies 4 \\lt 8k \\leq 184 \\implies \\dfrac{1}{2} \\lt k \\leq 23$$ +Так как $k$ целое, получаем $k = 1,\\ 2,\\ \\ldots,\\ 23$. +
Шаг 3. Найдём первый и последний члены: +$$a_1 = 8\\cdot 1 + 4 = 12, \\quad a_n = 8\\cdot 23 + 4 = 188$$ +Шаг 4. Числа $12,\\ 20,\\ 28,\\ \\ldots,\\ 188$ образуют арифметическую прогрессию с разностью $d = 8$. +
Шаг 5. Количество членов: $n = 23$. +
Шаг 6. Применим формулу суммы: +$$S = \\dfrac{n(a_1 + a_n)}{2} = \\dfrac{23\\cdot(12 + 188)}{2} = \\dfrac{23\\cdot 200}{2} = 23\\cdot 100 = 2300$$ +
Ответ: $2300$
` + }, + { + text: `Квадратный участок земли разбили на четыре части: газон, цветник, огород и сад. + Сад и цветник — квадраты. Периметр сада — $80$ м, а цветника — $20$ м. + Чему равен периметр газона?`, + sol: `Формула периметра квадрата: $P = 4a$, откуда сторона $a = \\dfrac{P}{4}$. +
Шаг 1. Находим стороны квадратных частей: +
— Сторона сада: $80 \\div 4 = 20$ м. +
— Сторона цветника: $20 \\div 4 = 5$ м. +
Шаг 2. Сторона всего квадратного участка $= 20 + 5 = 25$ м. + + + + + + + + + Газон + Цветник + Сад + Огород + 20 м + 5 м + 20 м + 20 м + 5 м + +Шаг 3. Газон — прямоугольник со сторонами $20$ м и $5$ м (по рисунку). +
По формуле периметра прямоугольника $P = 2(a + b)$: +$$P_{\\text{газон}} = 2(20 + 5) = 50\\text{ м}$$ +
Ответ: $50$ м
` + }, + { + text: `Дан треугольник $ABC$ со сторонами $AC = 9$, $BC = 12$, $AB = 15$. + Найдите расстояние между центрами описанной и вписанной окружностей треугольника.`, + sol: ` + + + + + + + + + + + + + + C + A + B + O + I + 4,5 + 6 + 3 + 3 + 1,5 + 3 + 3√5/2 + 9 + 12 + 15 + R = 7,5 + r = 3 + +Шаг 1. Определим тип треугольника с помощью обратной теоремы Пифагора: если $a^2 + b^2 = c^2$, то треугольник прямоугольный. +$$9^2 + 12^2 = 81 + 144 = 225 = 15^2 \\checkmark$$ +Значит, $\\triangle ABC$ — прямоугольный, причём прямой угол лежит против гипотенузы $AB = 15$, т.е. $\\angle C = 90°$. +
Шаг 2. Центр и радиус описанной окружности. +
Свойство: в прямоугольном треугольнике центр описанной окружности — это середина гипотенузы, а её радиус равен половине гипотенузы: +$$R = \\dfrac{AB}{2} = \\dfrac{15}{2} = 7{,}5\\text{ см}$$ +Шаг 3. Радиус вписанной окружности. +
Для прямоугольного треугольника с катетами $a$, $b$ и гипотенузой $c$: +$$r = \\dfrac{a + b - c}{2} = \\dfrac{9 + 12 - 15}{2} = \\dfrac{6}{2} = 3\\text{ см}$$ +Шаг 4. Координаты центров. +
Поместим $C = (0;\\,0)$, $A = (9;\\,0)$, $B = (0;\\,12)$. +
— Центр $O$ описанной окружности — середина $AB$: $O = \\left(\\dfrac{9}{2};\\,\\dfrac{12}{2}\\right) = (4{,}5;\\,6)$. +
— Центр $I$ вписанной окружности удалён на $r = 3$ от каждого катета, значит $I = (3;\\,3)$. +
Шаг 5. Расстояние между центрами по формуле расстояния между точками: +$$OI = \\sqrt{(4{,}5 - 3)^2 + (6 - 3)^2} = \\sqrt{1{,}5^2 + 3^2} = \\sqrt{2{,}25 + 9} = \\sqrt{11{,}25}$$ +$$\\sqrt{11{,}25} = \\sqrt{\\dfrac{45}{4}} = \\dfrac{\\sqrt{45}}{2} = \\dfrac{3\\sqrt{5}}{2}\\text{ см}$$ +
Ответ: $\\dfrac{3\\sqrt{5}}{2}$ см
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v37.js b/frontend/js/exam9/variants/v37.js new file mode 100644 index 0000000..04b29ce --- /dev/null +++ b/frontend/js/exam9/variants/v37.js @@ -0,0 +1,201 @@ +VARIANTS[37] = { + label: "Вариант 37", + tasks: [ + { + text: `Какое из данных равенств является верным, если $f(x) = \\dfrac{6}{x}$:`, + opts: [ + ["а", "$f(2) = 2$"], ["б", "$f(2) = 3$"], ["в", "$f(2) = 12$"], + ["г", "$f(2) = 4$"], ["д", "$f(2) = 36$"], + ], + sol: `Подставляем $x=2$: +$$f(2) = \\dfrac{6}{2} = 3$$ +
Ответ: б) $f(2)=3$
` + }, + { + text: `Значение выражения $\\sqrt{\\dfrac{25}{16}} - \\dfrac{1}{4}$ равно:`, + opts: [ + ["а", "$\\dfrac{1}{4}$"], ["б", "$\\sqrt{\\dfrac{21}{16}}$"], ["в", "$\\sqrt{\\dfrac{29}{16}}$"], + ["г", "$1$"], ["д", "$\\dfrac{3}{2}$"], + ], + sol: `$$\\sqrt{\\dfrac{25}{16}} = \\dfrac{\\sqrt{25}}{\\sqrt{16}} = \\dfrac{5}{4}$$ +$$\\dfrac{5}{4} - \\dfrac{1}{4} = \\dfrac{4}{4} = 1$$ +
Ответ: г) $1$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "если катет равен половине гипотенузы, то он лежит против угла в $30^{\\circ}$;"], + ["б", "если в треугольнике $ABC$ $R$ — радиус описанной окружности, то $\\dfrac{AC}{\\sin B} = 2R$;"], + ["в", "в любой квадрат можно вписать окружность;"], + ["г", "сумма углов любого треугольника равна $360^{\\circ}$?"], + ], + sol: `
    +
  • а) Катет $=$ половина гипотенузы ⟹ напротив угла $30°$ — верно
    • +
  • б) Теорема синусов $\\dfrac{AC}{\\sin B}=2R$ — верно
    • +
  • в) В любой квадрат вписывается окружность — верно
    • +
  • г) «Сумма углов треугольника $= 360°$» — НЕВЕРНО. Сумма углов любого треугольника равна $\\mathbf{180°}$.
    • +
+
Ответ: г)
` + }, + { + text: `Раскройте скобки $-10 - (-5x - 12)$ и упростите полученное выражение.`, + sol: `Минус перед скобками меняет знак каждого слагаемого внутри: +$$-10 - (-5x - 12) = -10 + 5x + 12 = 5x + 2$$ +
Ответ: $5x+2$
` + }, + { + text: `Решите неравенство $2x - 5 > 17$. + Определите количество целых чисел из второго десятка, которые являются решениями этого неравенства.`, + sol: `Свойства линейного неравенства: к обеим частям можно прибавлять одно и то же число, а также можно делить обе части на положительное число — знак неравенства не меняется. +
Шаг 1. Прибавляем $5$ к обеим частям: +$$2x - 5 + 5 > 17 + 5 \\implies 2x > 22$$ +Шаг 2. Делим обе части на $2$ (положительное число): +$$x > 11$$ +Шаг 3. Второй десяток — это натуральные числа от $11$ до $20$ (числа $11,\\,12,\\,\\ldots,\\,20$). +
Решениями (т.е. $x > 11$ строго) являются те, что больше $11$: +$$12,\\,13,\\,14,\\,15,\\,16,\\,17,\\,18,\\,19,\\,20$$ +Всего $9$ чисел. +
Ответ: $x>11$; целых из второго десятка — $9$
` + }, + { + text: `Периметр ромба $ABCD$ равен $48$ см, острый угол $A$ ромба равен $60^{\\circ}$. + Найдите меньшую диагональ $BD$ ромба.`, + sol: `Сторона ромба: $a = \\dfrac{48}{4} = 12$ см. + + + + + + + + + + + + A + B + C + D + 60° + + 12 + 12 + 12 + + BD=12 + +Ключевая идея. В ромбе $ABCD$ рассмотрим треугольник $ABD$: +
    +
  • $AB = AD = 12$ см (стороны ромба)
    • +
  • $\\angle BAD = 60°$ (по условию)
    • +
+Треугольник $ABD$ — равнобедренный с углом при вершине $60°$. Углы при основании равны $\\dfrac{180°-60°}{2}=60°$ — все три угла $= 60°$, т.е. треугольник равносторонний. +
Значит, $BD = AB = 12$ см. +
Ответ: $BD = 12$ см
` + }, + { + text: `В арифметической прогрессии второй и девятый члены соответственно равны $3$ и $10$. + Чему равна сумма третьего и десятого членов этой прогрессии?`, + sol: `Формула $n$-го члена арифметической прогрессии: $a_n = a_1 + (n - 1)d$, где $a_1$ — первый член, $d$ — разность прогрессии. +
Из этой формулы следует: $a_m - a_k = (m - k)\\cdot d$ для любых номеров $m$, $k$. +
Шаг 1. Находим разность $d$. +
Между $a_2$ и $a_9$ — $9 - 2 = 7$ шагов: +$$a_9 - a_2 = 7d \\implies 10 - 3 = 7d \\implies 7d = 7 \\implies d = 1$$ +Шаг 2. Находим $a_3$ и $a_{10}$. +
$a_3 = a_2 + d = 3 + 1 = 4$ (следующий член = предыдущий $+\\,d$). +
$a_{10} = a_9 + d = 10 + 1 = 11$. +
Шаг 3. Складываем: +$$a_3 + a_{10} = 4 + 11 = 15$$ +
Ответ: $15$
` + }, + { + text: `Двое сотрудников производственной лаборатории Слуцкого сыродельного комбината + $6$ дней обрабатывали результаты измерений. + За какое время может выполнить эту работу первый работник, работая отдельно, + если он за $2$ дня выполняет такую же часть работы, какую второй выполняет за $3$ дня?`, + sol: `Пусть производительности (часть работы за день): первый — $r_1$, второй — $r_2$. +
Условие 1: «первый за $2$ дня выполняет столько же, сколько второй за $3$»: +$$2r_1 = 3r_2 \\implies r_2 = \\dfrac{2}{3}r_1$$ +Условие 2: вместе за $6$ дней выполнили всю работу ($=1$): +$$6(r_1 + r_2) = 1 \\implies 6\\left(r_1 + \\dfrac{2}{3}r_1\\right) = 1$$ +$$6\\cdot\\dfrac{5r_1}{3} = 1 \\implies 10r_1 = 1 \\implies r_1 = \\dfrac{1}{10}$$ +Первый работник один выполнит работу за время $\\dfrac{1}{r_1} = 10$ дней. +
Ответ: $10$ дней
` + }, + { + text: `Окружности с радиусами $4$ см и $9$ см касаются внешним образом. + Найдите отрезок общей внешней касательной, заключённый между точками касания.`, + sol: `$O_1O_2 = R+r = 9+4 = 13$ см (внешнее касание). Радиусы перпендикулярны касательной: $O_1T_1\\perp T_1T_2$ и $O_2T_2\\perp T_1T_2$, поэтому $O_1T_1\\parallel O_2T_2$. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + O₁ + O₂ + T₁ + T₂ + H + + R=9 + r=4 + 5 + 12 + T₁T₂=? + 9 + 4 + +Ключевая идея: опустим из $O_2$ перпендикуляр на прямую $O_1T_1$ — получим точку $H$. +
Тогда $O_2T_2T_1H$ — прямоугольник, значит $O_2H = T_1T_2$ (то, что ищем). +$$O_1H = O_1T_1 - T_1H = R - r = 9 - 4 = 5\\text{ см}$$ +В прямоугольном $\\triangle O_1HO_2$ ($\\angle H = 90°$) — узнаём тройку 5–12–13: +$$O_1O_2^2 = O_1H^2 + O_2H^2 \\implies 13^2 = 5^2 + (T_1T_2)^2$$ +$$169 = 25 + (T_1T_2)^2 \\implies (T_1T_2)^2 = 144 \\implies T_1T_2 = 12\\text{ см}$$ +
Ответ: $12$ см
` + }, + { + text: `Определите знак выражения $8x_1 - x_2$, где $x_1$, $x_2$ — корни уравнения + $(6 + 2\\sqrt{5})\\,x^2 - 15x - (6 - 2\\sqrt{5}) = 0$ и $x_1 < x_2$.`, + sol: `Шаг 1. Замена переменной. +
Пусть $y = (6+2\\sqrt{5})\\,x$, тогда $x = \\dfrac{y}{6+2\\sqrt{5}}$. Подставляем в уравнение и умножаем на $(6+2\\sqrt{5})$: +$$y^2 - 15y - (6+2\\sqrt{5})(6-2\\sqrt{5}) = 0$$ +По формуле разности квадратов: $(6+2\\sqrt{5})(6-2\\sqrt{5}) = 36 - 20 = 16$. +$$y^2 - 15y - 16 = 0$$ +Шаг 2. Решаем простое уравнение. +$$D = 225 + 64 = 289 = 17^2 \\implies y = \\dfrac{15\\pm17}{2}$$ +$$y_1 = -1, \\quad y_2 = 16$$ +Шаг 3. Возвращаемся к $x$. $x = \\dfrac{y}{6+2\\sqrt{5}}$. +
Так как $6+2\\sqrt{5} > 0$, знак $x$ совпадает со знаком $y$. Поэтому $x_1 < 0 < x_2$: +$$x_1 = \\dfrac{-1}{6+2\\sqrt{5}} = -\\dfrac{6-2\\sqrt{5}}{16} \\quad\\text{(домножили на сопряжённое)}$$ +$$x_2 = \\dfrac{16}{6+2\\sqrt{5}} = \\dfrac{16(6-2\\sqrt{5})}{16} = 6-2\\sqrt{5}$$ +Шаг 4. Считаем $8x_1 - x_2$. +$$8x_1 = 8\\cdot\\left(-\\dfrac{6-2\\sqrt{5}}{16}\\right) = -\\dfrac{6-2\\sqrt{5}}{2}$$ +$$8x_1 - x_2 = -\\dfrac{6-2\\sqrt{5}}{2} - (6-2\\sqrt{5}) = -(6-2\\sqrt{5})\\left(\\dfrac{1}{2}+1\\right) = -\\dfrac{3(6-2\\sqrt{5})}{2}$$ +Так как $6 > 2\\sqrt{5}$ (потому что $36 > 20$), значит $6-2\\sqrt{5} > 0$, и всё выражение отрицательно. +
Ответ: $8x_1 - x_2 < 0$ (отрицательно)
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v38.js b/frontend/js/exam9/variants/v38.js new file mode 100644 index 0000000..f0252ff --- /dev/null +++ b/frontend/js/exam9/variants/v38.js @@ -0,0 +1,209 @@ +VARIANTS[38] = { + label: "Вариант 38", + tasks: [ + { + text: `Какое из данных равенств является верным, если $f(x) = \\dfrac{8}{x}$:`, + opts: [ + ["а", "$f(2) = 2$"], ["б", "$f(2) = 4$"], ["в", "$f(2) = 16$"], + ["г", "$f(2) = 6$"], ["д", "$f(2) = 64$"], + ], + sol: `Подставляем $x=2$: +$$f(2) = \\dfrac{8}{2} = 4$$ +
Ответ: б) $f(2)=4$
` + }, + { + text: `Значение выражения $\\sqrt{\\dfrac{9}{25}} - \\dfrac{1}{5}$ равно:`, + opts: [ + ["а", "$\\dfrac{3}{5}$"], ["б", "$\\sqrt{\\dfrac{14}{25}}$"], ["в", "$\\dfrac{4}{5}$"], + ["г", "$1$"], ["д", "$\\dfrac{2}{5}$"], + ], + sol: `$$\\sqrt{\\dfrac{9}{25}} = \\dfrac{\\sqrt{9}}{\\sqrt{25}} = \\dfrac{3}{5}$$ +$$\\dfrac{3}{5} - \\dfrac{1}{5} = \\dfrac{2}{5}$$ +
Ответ: д) $\\dfrac{2}{5}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "если катет лежит против угла в $30^{\\circ}$, то он равен половине гипотенузы;"], + ["б", "если в треугольнике $ABC$ $R$ — радиус описанной окружности, то $\\dfrac{BC}{\\sin A} = 2R$;"], + ["в", "сумма углов любого четырёхугольника равна $180^{\\circ}$;"], + ["г", "около любого квадрата можно описать окружность?"], + ], + sol: `
    +
  • а) Катет напротив угла $30°$ равен половине гипотенузы — верно
    • +
  • б) Теорема синусов $\\dfrac{BC}{\\sin A}=2R$ — верно
    • +
  • в) «Сумма углов четырёхугольника $=180°$» — НЕВЕРНО. Сумма углов любого четырёхугольника равна $360°$.
    • +
  • г) Около любого квадрата можно описать окружность — верно
    • +
+
Ответ: в)
` + }, + { + text: `Раскройте скобки $-4 - (5x + 3)$ и упростите полученное выражение.`, + sol: `Минус перед скобками меняет знак каждого слагаемого внутри: +$$-4 - (5x + 3) = -4 - 5x - 3 = -5x - 7$$ +
Ответ: $-5x-7$
` + }, + { + text: `Решите неравенство $5 + 2x > 7$. + Определите количество целых чисел из первого десятка, которые являются решениями этого неравенства.`, + sol: `Свойства линейного неравенства: из обеих частей можно вычитать одно и то же число, а также делить обе части на положительное число — знак неравенства не меняется. +
Шаг 1. Вычитаем $5$ из обеих частей: +$$5 + 2x - 5 > 7 - 5 \\implies 2x > 2$$ +Шаг 2. Делим обе части на $2$ (положительное число): +$$x > 1$$ +Шаг 3. Первый десяток — натуральные числа от $1$ до $10$ (числа $1,\\,2,\\,\\ldots,\\,10$). +
Решениями (т.е. $x > 1$ строго) являются те, что больше $1$: +$$2,\\,3,\\,4,\\,5,\\,6,\\,7,\\,8,\\,9,\\,10$$ +Всего $9$ чисел. +
Ответ: $x > 1$; целых из первого десятка — $9$
` + }, + { + text: `Меньшая диагональ $BD$ ромба $ABCD$ равна $10$ см, тупой угол $B$ ромба равен $120^{\\circ}$. + Найдите периметр ромба.`, + sol: ` + + + + + + + + + + + + + A + B + C + D + 60° + 120° + + BD=10 + +Свойства ромба: +
1) Все четыре стороны ромба равны. +
2) Диагонали ромба взаимно перпендикулярны и точкой пересечения делятся пополам. +
3) Диагонали ромба являются биссектрисами его углов. +
Шаг 1. Обозначим точку пересечения диагоналей через $O$. По свойству диагоналей: +$$BO = \\dfrac{BD}{2} = \\dfrac{10}{2} = 5\\text{ см}$$ +Шаг 2. Поскольку $BD$ — биссектриса тупого угла $B = 120°$: +$$\\angle OBC = \\dfrac{\\angle B}{2} = \\dfrac{120°}{2} = 60°$$ +Шаг 3. Рассмотрим прямоугольный треугольник $OBC$ (прямой угол при $O$). +
По определению косинуса: $\\cos\\alpha = \\dfrac{\\text{прилежащий катет}}{\\text{гипотенуза}}$. +
Прилежащий к углу $B$ катет — $BO$, гипотенуза — $BC$: +$$\\cos 60° = \\dfrac{BO}{BC} \\implies \\dfrac{1}{2} = \\dfrac{5}{BC} \\implies BC = 10\\text{ см}$$ +Шаг 4. Периметр ромба — это сумма всех сторон, а они равны: +$$P = 4\\cdot BC = 4\\cdot 10 = 40\\text{ см}$$ +
Ответ: $P = 40$ см
` + }, + { + text: `В арифметической прогрессии третий и десятый члены соответственно равны $12$ и $-2$. + Чему равна сумма второго и одиннадцатого членов этой прогрессии?`, + sol: `Формула $n$-го члена арифметической прогрессии: $a_n = a_1 + (n - 1)d$. +
Из неё следует: $a_m - a_k = (m - k)\\cdot d$. +
Шаг 1. Находим разность $d$. +
Между $a_3$ и $a_{10}$ — $10 - 3 = 7$ шагов: +$$a_{10} - a_3 = 7d \\implies -2 - 12 = 7d \\implies 7d = -14 \\implies d = -2$$ +Шаг 2. Находим $a_2$ и $a_{11}$. +
$a_2 = a_3 - d = 12 - (-2) = 14$ (предыдущий член = следующий $-\\,d$). +
$a_{11} = a_{10} + d = -2 + (-2) = -4$. +
Шаг 3. Складываем: +$$a_2 + a_{11} = 14 + (-4) = 10$$ +
Ответ: $10$
` + }, + { + text: `Двое сотрудников по озеленению Национального историко-культурного + музея-заповедника «Несвиж» могут выполнить уборку части парка за $12$ дней. + За сколько дней, работая отдельно, выполнит эту работу первый сотрудник, + если он за $2$ дня выполняет такую же часть работы, какую второй выполняет за $3$ дня?`, + sol: `Пусть производительности (часть работы за день): первый — $r_1$, второй — $r_2$. +
Условие 1: «первый за $2$ дня выполняет столько же, сколько второй за $3$»: +$$2r_1 = 3r_2 \\implies r_2 = \\dfrac{2}{3}r_1$$ +Условие 2: вместе за $12$ дней выполнили всю работу ($=1$): +$$12(r_1 + r_2) = 1 \\implies 12\\left(r_1 + \\dfrac{2}{3}r_1\\right) = 1$$ +$$12\\cdot\\dfrac{5r_1}{3} = 1 \\implies 20r_1 = 1 \\implies r_1 = \\dfrac{1}{20}$$ +Первый сотрудник один выполнит работу за $\\dfrac{1}{r_1} = 20$ дней. +
Ответ: $20$ дней
` + }, + { + text: `Окружности с радиусами $9$ см и $16$ см касаются внешним образом. + Найдите отрезок общей внешней касательной, заключённый между точками касания.`, + sol: `$O_1O_2 = R+r = 16+9 = 25$ см (внешнее касание). Радиусы перпендикулярны касательной: $O_1T_1\\perp T_1T_2$ и $O_2T_2\\perp T_1T_2$, поэтому $O_1T_1\\parallel O_2T_2$. + + + + + + + + + + + + + + + + + + + + + + + + + + + + O₁ + O₂ + T₁ + T₂ + H + + R=16 + r=9 + 7 + 24 + T₁T₂=? + 16 + 9 + +Опустим из $O_2$ перпендикуляр на прямую $O_1T_1$ — получим точку $H$. +
Тогда $O_2T_2T_1H$ — прямоугольник, значит $O_2H = T_1T_2$ (то, что ищем). +$$O_1H = O_1T_1 - T_1H = R - r = 16 - 9 = 7\\text{ см}$$ +В прямоугольном $\\triangle O_1HO_2$ ($\\angle H = 90°$) — узнаём тройку 7–24–25: +$$O_1O_2^2 = O_1H^2 + O_2H^2 \\implies 25^2 = 7^2 + (T_1T_2)^2$$ +$$625 = 49 + (T_1T_2)^2 \\implies (T_1T_2)^2 = 576 \\implies T_1T_2 = 24\\text{ см}$$ +
Ответ: $24$ см
` + }, + { + text: `Определите знак выражения $\\dfrac{2}{9}x_1 - x_2$, где $x_1$, $x_2$ — корни уравнения + $(5 - 2\\sqrt{6})\\,x^2 - 10x + 9(5 + 2\\sqrt{6}) = 0$ и $x_1 \\gt x_2$.`, + sol: `Формула разности квадратов: $(a - b)(a + b) = a^2 - b^2$. +
Шаг 1. Подготовка. +
Замечаем красивое равенство: +$$(5 - 2\\sqrt{6})(5 + 2\\sqrt{6}) = 5^2 - (2\\sqrt{6})^2 = 25 - 24 = 1$$ +Числа $5 - 2\\sqrt{6}$ и $5 + 2\\sqrt{6}$ — взаимно обратны: $\\dfrac{1}{5 - 2\\sqrt{6}} = 5 + 2\\sqrt{6}$. +
Шаг 2. Замена переменной. +
Сделаем замену $y = (5 - 2\\sqrt{6})\\,x$. Тогда $x = y\\cdot(5 + 2\\sqrt{6})$. +
Подставляем в исходное уравнение, помня, что $(5 - 2\\sqrt{6})\\cdot x^2 = \\dfrac{y^2}{5 - 2\\sqrt{6}} = y^2(5 + 2\\sqrt{6})$... Удобнее иначе: умножим обе части исходного уравнения на $(5 - 2\\sqrt{6})$. Учитывая, что $9(5 + 2\\sqrt{6})\\cdot(5 - 2\\sqrt{6}) = 9\\cdot 1 = 9$: +$$(5 - 2\\sqrt{6})^2 x^2 - 10(5 - 2\\sqrt{6})\\,x + 9 = 0$$ +В переменной $y = (5 - 2\\sqrt{6})\\,x$ это: +$$y^2 - 10y + 9 = 0$$ +Шаг 3. Решаем по теореме Виета. +
Корни уравнения $y^2 - 10y + 9 = 0$: ищем числа с суммой $10$ и произведением $9$. Это $1$ и $9$: +$$(y - 1)(y - 9) = 0 \\implies y_1 = 1,\\; y_2 = 9$$ +Шаг 4. Возвращаемся к $x$. +
$x = y\\cdot(5 + 2\\sqrt{6})$, причём $5 + 2\\sqrt{6} \\gt 0$. По условию $x_1 \\gt x_2$, поэтому большему $y$ соответствует больший $x$: +$$x_1 = 9(5 + 2\\sqrt{6}), \\quad x_2 = 5 + 2\\sqrt{6}$$ +Шаг 5. Вычисляем $\\dfrac{2}{9}x_1 - x_2$. +$$\\dfrac{2}{9}\\cdot 9(5 + 2\\sqrt{6}) - (5 + 2\\sqrt{6}) = 2(5 + 2\\sqrt{6}) - (5 + 2\\sqrt{6}) = 5 + 2\\sqrt{6}$$ +Так как $5 \\gt 0$ и $2\\sqrt{6} \\gt 0$, то $5 + 2\\sqrt{6} \\gt 0$. +
Ответ: $\\dfrac{2}{9}x_1 - x_2 \\gt 0$ (положительно)
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v39.js b/frontend/js/exam9/variants/v39.js new file mode 100644 index 0000000..84167e8 --- /dev/null +++ b/frontend/js/exam9/variants/v39.js @@ -0,0 +1,216 @@ +VARIANTS[39] = { + label: "Вариант 39", + tasks: [ + { + text: `Определите промежуток, которому принадлежит число $1{,}15$:`, + opts: [ + ["а", "$(1{,}1;\\; 1{,}145)$"], ["б", "$(1;\\; 1{,}05)$"], ["в", "$(1;\\; 1{,}1)$"], + ["г", "$(1{,}2;\\; 1{,}3)$"], ["д", "$(1{,}1;\\; 1{,}2)$"], + ], + sol: `Проверяем каждый вариант: $1{,}1 < 1{,}15 < 1{,}2$ ✓ +
Ответ: д) $(1{,}1;\\;1{,}2)$
` + }, + { + text: `Значение выражения $4{,}5 : 9 - 0{,}4$ равно:`, + opts: [ + ["а", "$0{,}9$"], ["б", "$4{,}6$"], ["в", "$0{,}1$"], ["г", "$0{,}3$"], ["д", "$0{,}01$"], + ], + sol: `$$4{,}5 : 9 - 0{,}4 = 0{,}5 - 0{,}4 = 0{,}1$$ +
Ответ: в) $0{,}1$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "если все стороны квадрата увеличить в $2$ раза, то его площадь увеличится в $2$ раза;"], + ["б", "внешний угол треугольника является смежным с его внутренним углом;"], + ["в", "медианы равностороннего треугольника равны между собой;"], + ["г", "диагональ квадрата со стороной $a$ равна $a\\sqrt{2}$?"], + ], + sol: `
    +
  • а) Стороны $\\times2$ ⟹ площадь $\\times 2^2 = 4$, а не в $2$ — НЕВЕРНО
    • +
  • б) Внешний угол смежен с внутренним — верно
    • +
  • в) В равностороннем треугольнике все медианы равны — верно
    • +
  • г) Диагональ квадрата $= a\\sqrt{2}$ — верно
    • +
+
Ответ: а)
` + }, + { + text: `Выполните деление: $\\dfrac{m}{m+1} : \\dfrac{m}{m^2-1}$.`, + sol: `Деление на дробь — умножаем на обратную. ОДЗ: $m\\neq0$, $m\\neq\\pm1$. +$$\\dfrac{m}{m+1}:\\dfrac{m}{m^2-1} = \\dfrac{m}{m+1}\\cdot\\dfrac{m^2-1}{m} = \\dfrac{\\cancel{m}}{m+1}\\cdot\\dfrac{(m-1)\\cancel{(m+1)}}{\\cancel{m}} = m-1$$ +
Ответ: $m-1$
` + }, + { + text: `Найдите больший корень уравнения $x^4 - 15x^2 - 16 = 0$.`, + sol: `Метод решения биквадратного уравнения: уравнение вида $ax^4 + bx^2 + c = 0$ решается заменой $t = x^2$, причём $t \\geq 0$ (так как квадрат любого числа неотрицателен). +
Шаг 1. Делаем замену $t = x^2$, где $t \\geq 0$: +$$t^2 - 15t - 16 = 0$$ +Шаг 2. Решаем квадратное уравнение по формуле дискриминанта $D = b^2 - 4ac$: +$$D = (-15)^2 - 4\\cdot 1\\cdot(-16) = 225 + 64 = 289 = 17^2$$ +$$t = \\dfrac{15 \\pm 17}{2} \\implies t_1 = \\dfrac{32}{2} = 16, \\quad t_2 = \\dfrac{-2}{2} = -1$$ +Шаг 3. Поскольку $t = x^2 \\geq 0$, корень $t_2 = -1$ не подходит. Остаётся $t = 16$. +
Шаг 4. Возвращаемся к $x$: +$$x^2 = 16 \\implies x = \\pm\\sqrt{16} = \\pm 4$$ +Шаг 5. Из корней $x = 4$ и $x = -4$ больший — это $x = 4$. +
Ответ: больший корень $x = 4$
` + }, + { + text: `В треугольнике $ABC$ $AB = 6$ см, $AC = 5$ см, $CM$ — медиана, + $\\angle ACM = \\angle BCM$. Найдите синус угла $A$.`, + sol: `Шаг 1. Определяем тип треугольника. +
$CM$ — медиана, значит $M$ — середина $AB$. Равенство $\\angle ACM=\\angle BCM$ означает, что $CM$ — также биссектриса угла $C$. +
По теореме о биссектрисе: $\\dfrac{AM}{MB} = \\dfrac{AC}{BC}$. Так как $AM=MB$ (медиана), то $AC = BC$. +$$BC = AC = 5\\text{ см}$$ +Треугольник равнобедренный с основанием $AB=6$ и боковыми сторонами $=5$. + + + + + + A + B + C + M + 5 + 5 + 6 + A + h + +Шаг 2. Находим высоту $CM$. +
В равнобедренном треугольнике медиана из $C$ на $AB$ является одновременно высотой. Из прямоугольного $\\triangle CAM$: +$$h = CM = \\sqrt{AC^2 - AM^2} = \\sqrt{25 - 9} = \\sqrt{16} = 4\\text{ см}$$ +Шаг 3. Синус угла $A$. +$$\\sin A = \\dfrac{h}{AC} = \\dfrac{4}{5}$$ +
Ответ: $\\sin A = \\dfrac{4}{5}$
` + }, + { + text: `Найдите число целых решений неравенства + $\\dfrac{(x-3)(-x^2+5x+6)}{x-5} \\geq 0$.`, + sol: `Раскладываем числитель: +$$-x^2+5x+6 = -(x^2-5x-6) = -(x-6)(x+1)$$ +Неравенство принимает вид: +$$\\dfrac{-(x-3)(x-6)(x+1)}{x-5}\\geq0 \\iff \\dfrac{(x-3)(x-6)(x+1)}{x-5}\\leq0$$ +Метод интервалов. Корни: $x=-1,\\,3,\\,5,\\,6$ (при $x=5$ — ОДЗ). + + + + + + + + + + + +
ИнтервалЗнак$\\leq0$?
$x<-1$$+$✗
$x=-1$$0$✓
$-1\\lt x\\lt 3$$-$✓
$x=3$$0$✓
$3\\lt x\\lt 5$$+$✗
$x=5$не определено
$5\\lt x\\lt 6$$-$✓
$x=6$$0$✓
$x>6$$+$✗
+Решение: $x\\in[-1;\\,3]\\cup(5;\\,6]$. + + + + −1 + 3 + 5 + 6 + + + + + + + +Целые из $[-1;3]$: $-1,0,1,2,3$ — $5$ чисел. Из $(5;6]$: $6$ — $1$ число. +
Ответ: $6$ целых решений
` + }, + { + text: `График линейной функции проходит через точки $A(-2;\\; 11)$ и $B(4;\\; -10)$. + Запишите формулу, задающую эту функцию. + Найдите, при каких значениях переменной функция принимает неположительные значения.`, + sol: `Линейная функция: график $y = kx + b$ — прямая. Чтобы найти $k$ и $b$, подставляем координаты двух точек графика. +
Шаг 1. По условию график проходит через $A(-2;\\,11)$ и $B(4;\\,-10)$. Угловой коэффициент по формуле $k = \\dfrac{y_2 - y_1}{x_2 - x_1}$: +$$k = \\dfrac{-10 - 11}{4 - (-2)} = \\dfrac{-21}{6} = -\\dfrac{7}{2}$$ +Шаг 2. Найдём $b$, подставив координаты точки $A(-2;\\,11)$ в уравнение $y = kx + b$: +$$11 = -\\dfrac{7}{2}\\cdot(-2) + b \\implies 11 = 7 + b \\implies b = 4$$ +Значит, формула функции: $f(x) = -\\dfrac{7}{2}x + 4$. +
Шаг 3. Проверка: подставим $x = 4$: +$$f(4) = -\\dfrac{7}{2}\\cdot 4 + 4 = -14 + 4 = -10 \\quad \\checkmark$$ +Шаг 4. «Функция принимает неположительные значения» означает $f(x) \\leq 0$. Решаем неравенство: +$$-\\dfrac{7}{2}x + 4 \\leq 0$$ +Переносим $4$ в правую часть со сменой знака: +$$-\\dfrac{7}{2}x \\leq -4$$ +Делим на $-\\dfrac{7}{2}$ (отрицательное число — знак неравенства меняется на противоположный): +$$x \\geq \\dfrac{-4}{-7/2} = \\dfrac{8}{7}$$ +
Ответ: $f(x)=-\\dfrac{7}{2}x+4$; функция неположительна при $x\\geq\\dfrac{8}{7}$
` + }, + { + text: `Найдите площадь треугольника, если две его стороны равны $6$ см и $8$ см, + а медиана, проведённая к третьей стороне, равна $5$ см.`, + sol: `Шаг 1. Строим параллелограмм. +
Пусть $M$ — середина третьей стороны $AB$, а $CM=5$ — медиана. Отметим точку $D$ так, чтобы $M$ стала серединой отрезка $CD$ (то есть $MD=CM=5$, $CD=10$). +
Тогда $ACBD$ — параллелограмм, ведь его диагонали $AB$ и $CD$ делятся точкой $M$ пополам. + + + + + + + + + + + + + + + + + C + A + B + D + M + + 6 + 8 + 8 + 6 + 5 + 5 + 10 + +Шаг 2. Стороны параллелограмма. +
В параллелограмме $ACBD$: $AC = BD = 6$ и $BC = AD = 8$ (противоположные стороны). Диагональ $CD = 2\\cdot CM = 10$. +
Шаг 3. Треугольник $ACD$ — прямоугольный. +
Рассмотрим $\\triangle ACD$ со сторонами $AC=6$, $AD=8$, $CD=10$: +$$6^2 + 8^2 = 36 + 64 = 100 = 10^2 \\checkmark$$ +По обратной теореме Пифагора: $\\angle A = 90°$ (зелёный прямой угол на рисунке). +$$S_{\\triangle ACD} = \\dfrac{1}{2}\\cdot AC\\cdot AD = \\dfrac{1}{2}\\cdot6\\cdot8 = 24\\text{ см}^2$$ +Шаг 4. Площадь исходного треугольника. +
Диагональ $CD$ делит параллелограмм на два равных треугольника: $\\triangle ACD$ и $\\triangle ABC$. +$$S_{\\triangle ABC} = S_{\\triangle ACD} = 24\\text{ см}^2$$ +
Ответ: $24$ см²
` + }, + { + text: `Из двух домов, расстояние между которыми $180$ м, вышли и одновременно пошли + в одном направлении в школу мальчик и девочка. Девочка идёт впереди мальчика. + Скорость мальчика $6$ км/ч, скорость девочки $60$ м/мин. + Догонит ли мальчик девочку до прихода в школу, если путь девочки занимает $4$ мин? + Ответ обоснуйте.`, + sol: `Переводим скорость мальчика: +$$6\\text{ км/ч} = \\dfrac{6000\\text{ м}}{60\\text{ мин}} = 100\\text{ м/мин}$$ +Расстояние до школы (от девочки): $60\\times4=240$ м. Мальчик стартует на $180$ м позади, значит ему до школы $240+180=420$ м. +
Скорость сближения: мальчик быстрее на $100-60=40$ м/мин. Начальный разрыв $=180$ м. +
Время до нагона: +$$t = \\dfrac{180}{40} = 4{,}5\\text{ мин}$$ +Но девочка добирается до школы за $4$ мин, а мальчику нужно $4{,}5>4$ мин, чтобы её нагнать. +
Проверим по позициям (отсчёт от старта девочки): + + + + +
МоментДевочкаМальчик
$t=0$$0$ м$-180$ м
$t=4$ мин$240$ м (школа) ✓$-180+400=220$ м
+В момент, когда девочка прибыла в школу ($240$ м), мальчик находится на расстоянии $240-220=20$ м позади. +
Ответ: нет, мальчик не догонит — время нагона $4{,}5$ мин, а путь девочки $4$ мин
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v40.js b/frontend/js/exam9/variants/v40.js new file mode 100644 index 0000000..047e3c0 --- /dev/null +++ b/frontend/js/exam9/variants/v40.js @@ -0,0 +1,202 @@ +VARIANTS[40] = { + label: "Вариант 40", + tasks: [ + { + text: `Определите промежуток, которому принадлежит число $2{,}21$:`, + opts: [ + ["а", "$(2{,}1;\\; 2{,}121)$"], ["б", "$(2;\\; 2{,}01)$"], ["в", "$(2;\\; 2{,}1)$"], + ["г", "$(2{,}2;\\; 2{,}3)$"], ["д", "$(2{,}1;\\; 2{,}2)$"], + ], + sol: `Проверяем каждый вариант: $2{,}2 < 2{,}21 < 2{,}3$ ✓ +
Ответ: г) $(2{,}2;\\;2{,}3)$
` + }, + { + text: `Значение выражения $-1{,}5 : 3 - 0{,}4$ равно:`, + opts: [ + ["а", "$-0{,}9$"], ["б", "$-0{,}1$"], ["в", "$0{,}1$"], ["г", "$0{,}9$"], ["д", "$-5{,}4$"], + ], + sol: `$$-1{,}5 : 3 - 0{,}4 = -0{,}5 - 0{,}4 = -0{,}9$$ +
Ответ: а) $-0{,}9$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "если все стороны квадрата уменьшить в $2$ раза, то его площадь уменьшится в $2$ раза;"], + ["б", "в треугольнике против большего угла лежит большая сторона;"], + ["в", "сторона квадрата с диагональю $d$ равна $\\dfrac{d}{\\sqrt{2}}$;"], + ["г", "внешний угол треугольника равен сумме двух внутренних углов, не смежных с ним?"], + ], + sol: `
    +
  • а) Стороны $\\div2$ ⟹ площадь $\\div 2^2 = 4$, а не в $2$ — НЕВЕРНО
    • +
  • б) В треугольнике против большего угла лежит большая сторона — верно
    • +
  • в) Диагональ $d = a\\sqrt{2} \\Rightarrow a = \\dfrac{d}{\\sqrt{2}}$ — верно
    • +
  • г) Внешний угол равен сумме двух несмежных внутренних — верно
    • +
+
Ответ: а)
` + }, + { + text: `Выполните деление: $\\dfrac{m^2-1}{m} : \\dfrac{m-1}{m}$.`, + sol: `Деление на дробь — умножаем на обратную. ОДЗ: $m\\neq0$, $m\\neq1$. +$$\\dfrac{m^2-1}{m}:\\dfrac{m-1}{m} = \\dfrac{m^2-1}{m}\\cdot\\dfrac{m}{m-1} = \\dfrac{(m-1)(m+1)}{\\cancel{m}}\\cdot\\dfrac{\\cancel{m}}{m-1} = m+1$$ +
Ответ: $m+1$
` + }, + { + text: `Найдите больший корень уравнения $x^4 - 8x^2 - 9 = 0$.`, + sol: `Метод решения биквадратного уравнения: уравнение вида $ax^4 + bx^2 + c = 0$ решается заменой $t = x^2$, где $t \\geq 0$ (квадрат числа неотрицателен). +
Шаг 1. Делаем замену $t = x^2$, $t \\geq 0$: +$$t^2 - 8t - 9 = 0$$ +Шаг 2. Решаем по теореме Виета (ищем числа с суммой $8$ и произведением $-9$): это $9$ и $-1$. +$$(t - 9)(t + 1) = 0 \\implies t_1 = 9,\\; t_2 = -1$$ +Шаг 3. Корень $t_2 = -1$ не подходит, так как $t = x^2 \\geq 0$. Остаётся $t = 9$. +
Шаг 4. Возвращаемся к $x$: +$$x^2 = 9 \\implies x = \\pm\\sqrt{9} = \\pm 3$$ +Шаг 5. Больший из корней $\\{-3,\\,3\\}$ — это $x = 3$. +
Ответ: больший корень $x = 3$
` + }, + { + text: `В треугольнике $ABC$ $BC = 10$ см, $CM$ — биссектриса, $AM = MB = 8$ см. + Найдите синус угла $B$.`, + sol: `Свойство: если в треугольнике биссектриса является одновременно медианой, то треугольник — равнобедренный (с боковыми сторонами, выходящими из этой вершины). +
Шаг 1. Условие $AM = MB = 8$ см означает, что $M$ — середина $AB$, то есть $CM$ является медианой из вершины $C$. По условию $CM$ — также биссектриса. +
По указанному свойству $\\triangle ABC$ равнобедренный с $AC = BC$, причём $BC = 10$ см дано, значит $AC = 10$ см. +
$AB = AM + MB = 8 + 8 = 16$ см. + + + + + + A + B + C + M + 10 + 10 + 16 + h + +
Шаг 2. Применим теорему косинусов: $c^2 = a^2 + b^2 - 2ab\\cos\\gamma$, где $\\gamma$ — угол между сторонами $a$ и $b$, $c$ — сторона напротив этого угла. +
В $\\triangle ABC$: сторона $AC$ лежит напротив угла $B$, $BC$ и $AB$ — стороны, выходящие из вершины $B$. Поэтому: +$$AC^2 = BC^2 + AB^2 - 2\\cdot BC\\cdot AB\\cdot\\cos B$$ +$$10^2 = 10^2 + 16^2 - 2\\cdot 10\\cdot 16\\cdot\\cos B$$ +$$100 = 100 + 256 - 320\\cos B$$ +$$320\\cos B = 256 \\implies \\cos B = \\dfrac{256}{320} = \\dfrac{4}{5}$$ +Шаг 3. Применим основное тригонометрическое тождество: $\\sin^2\\alpha + \\cos^2\\alpha = 1$. +
Угол $B$ — острый (так как $\\cos B \\gt 0$), значит $\\sin B \\gt 0$: +$$\\sin B = \\sqrt{1 - \\cos^2 B} = \\sqrt{1 - \\dfrac{16}{25}} = \\sqrt{\\dfrac{9}{25}} = \\dfrac{3}{5}$$ +
Ответ: $\\sin B = \\dfrac{3}{5}$
` + }, + { + text: `Найдите число целых решений неравенства + $\\dfrac{(x+3)(-x^2+3x+4)}{x+2} \\geq 0$.`, + sol: `Метод интервалов применяется для решения рациональных неравенств: находим корни числителя и знаменателя, отмечаем их на числовой прямой, определяем знак выражения на каждом интервале. +
Шаг 1. Разложим квадратный трёхчлен в скобках. По теореме Виета ($x_1 + x_2 = 3$, $x_1 x_2 = -4$): корни $4$ и $-1$. +$$-x^2 + 3x + 4 = -(x^2 - 3x - 4) = -(x - 4)(x + 1)$$ +Шаг 2. Подставляем в неравенство: +$$\\dfrac{(x + 3)\\cdot[-(x - 4)(x + 1)]}{x + 2} \\geq 0$$ +Умножим обе части на $-1$ (знак неравенства меняется!): +$$\\dfrac{(x + 3)(x - 4)(x + 1)}{x + 2} \\leq 0$$ +Шаг 3. Находим корни числителя ($-3,\\,-1,\\,4$) и точку, в которой выражение не определено: $x = -2$ (знаменатель равен нулю). +
Шаг 4. Расставляем точки на числовой прямой и определяем знаки методом интервалов. Получаем: +$$x \\in [-3;\\,-2)\\cup[-1;\\,4]$$ +(в точке $x = -2$ знаменатель обнуляется, поэтому она исключается; остальные корни числителя включаются, так как неравенство нестрогое). + + + + −3 + −2 + −1 + 4 + + + + + + + +Шаг 5. Считаем целые числа в найденных интервалах: +
— Из $[-3;\\,-2)$: только $-3$ (1 число). +
— Из $[-1;\\,4]$: $-1, 0, 1, 2, 3, 4$ (6 чисел). +
Всего: $1 + 6 = 7$. +
Ответ: $7$ целых решений
` + }, + { + text: `График линейной функции проходит через точки $A(-2;\\; 1)$ и $B(-1;\\; -3)$. + Запишите формулу, задающую эту функцию. + Найдите, при каких значениях переменной функция принимает неотрицательные значения.`, + sol: `Линейная функция: график $y = kx + b$ — прямая; коэффициенты $k$ и $b$ находятся подстановкой координат двух точек графика. +
Шаг 1. Найдём угловой коэффициент по формуле $k = \\dfrac{y_2 - y_1}{x_2 - x_1}$: +$$k = \\dfrac{-3 - 1}{-1 - (-2)} = \\dfrac{-4}{1} = -4$$ +Шаг 2. Найдём $b$, подставив точку $A(-2;\\,1)$ в уравнение $y = kx + b$: +$$1 = -4\\cdot(-2) + b \\implies 1 = 8 + b \\implies b = -7$$ +Значит, $f(x) = -4x - 7$. +
Шаг 3. Проверка по второй точке $B(-1;\\,-3)$: +$$f(-1) = -4\\cdot(-1) - 7 = 4 - 7 = -3 \\quad \\checkmark$$ +Шаг 4. «Функция принимает неотрицательные значения» означает $f(x) \\geq 0$. Решаем неравенство: +$$-4x - 7 \\geq 0 \\implies -4x \\geq 7$$ +Делим обе части на $-4$ (отрицательное число — знак неравенства меняется): +$$x \\leq -\\dfrac{7}{4}$$ +
Ответ: $f(x)=-4x-7$; функция неотрицательна при $x\\leq-\\dfrac{7}{4}$
` + }, + { + text: `Найдите площадь треугольника, если две его стороны равны $5$ см и $12$ см, + а медиана, проведённая к третьей стороне, равна $6{,}5$ см.`, + sol: `Шаг 1. Строим параллелограмм. +
Пусть $M$ — середина третьей стороны $AB$, а $CM=6{,}5$ — медиана. Отметим точку $D$ так, чтобы $M$ стала серединой отрезка $CD$ ($MD=CM=6{,}5$, $CD=13$). +
Тогда $ACBD$ — параллелограмм, диагонали $AB$ и $CD$ делятся точкой $M$ пополам. + + + + + + + + + C + A + B + D + M + 5 + 12 + 12 + 5 + 6,5 + 6,5 + 13 + +Шаг 2. Стороны параллелограмма. +
$AC=BD=5$, $BC=AD=12$, диагональ $CD=2\\cdot6{,}5=13$. +Шаг 3. Треугольник $ACD$ — прямоугольный. +
Стороны $AC=5$, $AD=12$, $CD=13$: +$$5^2 + 12^2 = 25 + 144 = 169 = 13^2 \\checkmark$$ +По обратной теореме Пифагора: $\\angle A = 90°$. +$$S_{\\triangle ACD} = \\dfrac{1}{2}\\cdot AC\\cdot AD = \\dfrac{1}{2}\\cdot5\\cdot12 = 30\\text{ см}^2$$ +Шаг 4. Площадь исходного треугольника. +
Диагональ $CD$ делит параллелограмм на два равных треугольника: +$$S_{\\triangle ABC} = S_{\\triangle ACD} = 30\\text{ см}^2$$ +
Ответ: $30$ см²
` + }, + { + text: `Из двух домов, расстояние между которыми $280$ м, вышли и одновременно пошли + в одном направлении в школу мальчик и девочка. Девочка идёт впереди мальчика. + Скорость девочки $70$ м/мин, скорость мальчика $5{,}4$ км/ч. + Догонит ли мальчик девочку до прихода в школу, если путь девочки занимает $6$ мин? + Ответ обоснуйте.`, + sol: `Переводим скорость мальчика: +$$5{,}4\\text{ км/ч} = \\dfrac{5400\\text{ м}}{60\\text{ мин}} = 90\\text{ м/мин}$$ +Расстояние до школы (от девочки): $70\\times6=420$ м. Мальчик стартует на $280$ м позади, значит ему до школы $420+280=700$ м. +
Скорость сближения: мальчик быстрее на $90-70=20$ м/мин. Начальный разрыв $=280$ м. +
Время до нагона: +$$t = \\dfrac{280}{20} = 14\\text{ мин}$$ +Но девочка добирается до школы за $6$ мин, а мальчику нужно $14>6$ мин, чтобы её нагнать. +
Проверим по позициям (отсчёт от старта девочки): + + + + +
МоментДевочкаМальчик
$t=0$$0$ м$-280$ м
$t=6$ мин$420$ м (школа) ✓$-280+540=260$ м
+В момент, когда девочка прибыла в школу ($420$ м), мальчик находится на расстоянии $420-260=160$ м позади. +
Ответ: нет, мальчик не догонит — время нагона $14$ мин, а путь девочки $6$ мин
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v41.js b/frontend/js/exam9/variants/v41.js new file mode 100644 index 0000000..43918f3 --- /dev/null +++ b/frontend/js/exam9/variants/v41.js @@ -0,0 +1,186 @@ +VARIANTS[41] = { + label: "Вариант 41", + tasks: [ + { + text: `Определите верное равенство:`, + opts: [ + ["а", "$7\\% = 70$"], ["б", "$7\\% = 7$"], ["в", "$7\\% = 0{,}7$"], + ["г", "$7\\% = 0{,}07$"], ["д", "$7\\% = 0{,}007$"], + ], + sol: `$7\\% = \\dfrac{7}{100} = 0{,}07$ +
Ответ: г) $7\\%=0{,}07$
` + }, + { + text: `Запишите одночлен $3a^4b \\cdot \\left(-\\dfrac{1}{3}a^3\\right)$ в стандартном виде:`, + opts: [ + ["а", "$-ab$"], ["б", "$-a^7b$"], ["в", "$a^4b$"], + ["г", "$2\\dfrac{2}{3}ab$"], ["д", "$9a^7b$"], + ], + sol: `$$3a^4b\\cdot\\left(-\\dfrac{1}{3}a^3\\right) = \\left(3\\cdot\\left(-\\dfrac{1}{3}\\right)\\right)\\cdot a^{4+3}\\cdot b = -a^7b$$ +
Ответ: б) $-a^7b$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "радиус описанной окружности прямоугольного треугольника с гипотенузой $c$ находится по формуле $R = \\dfrac{c}{2}$;"], + ["б", "площадь прямоугольного треугольника с катетами $a$ и $b$ находится по формуле $S = \\dfrac{ab}{2}$;"], + ["в", "расстояние от точки до прямой определяется длиной перпендикуляра, проведённого к этой прямой;"], + ["г", "если в треугольнике $ABC$ сторона $BC$ — наибольшая, то угол $C$ — наибольший?"], + ], + sol: `
    +
  • а) $R=c/2$ для прямоугольного треугольника — верно
    • +
  • б) $S=ab/2$ — верно
    • +
  • в) Расстояние = длина перпендикуляра — верно
    • +
  • г) Сторона $BC$ лежит напротив угла $A$. Наибольшая сторона ⟹ наибольший противолежащий угол. Значит наибольший угол — $\\angle A$, а не $\\angle C$ — НЕВЕРНО
    • +
+
Ответ: г)
` + }, + { + text: `График функции $y = kx - 4$ проходит через точку $K\\!\\left(-\\dfrac{1}{2};\\; 4\\right)$. + Найдите коэффициент $k$.`, + sol: `Подставляем координаты точки $K$ в уравнение: +$$4 = k\\cdot\\left(-\\dfrac{1}{2}\\right) - 4 \\implies 8 = -\\dfrac{k}{2} \\implies k = -16$$ +
Ответ: $k = -16$
` + }, + { + text: `Диагонали параллелограмма $ABCD$ пересекаются в точке $O$, + площадь треугольника $AOB$ равна $15$ см². + Высота, проведённая из вершины $C$ к $AD$, равна $6$ см. + Найдите длину стороны $BC$ параллелограмма.`, + sol: `Шаг 1. Площадь параллелограмма. +
Диагонали делят параллелограмм на $4$ равновеликих треугольника: +$$S_{ABCD} = 4\\cdot S_{AOB} = 4\\cdot15 = 60\\text{ см}^2$$ + + + + + + + + + A + D + B + C + O + 15 + h=6 + +Шаг 2. Длина $AD$. +
Высота из $C$ к $AD$ — это расстояние между параллельными сторонами $BC$ и $AD$ (высота параллелограмма): +$$S_{ABCD} = AD\\cdot h \\implies 60 = AD\\cdot6 \\implies AD = 10\\text{ см}$$ +Шаг 3. В параллелограмме $BC\\parallel AD$, поэтому $BC = AD = 10$ см. +
Ответ: $BC = 10$ см
` + }, + { + text: `Определите, при каких значениях переменной разность дробей + $\\dfrac{4a-3}{5}$ и $\\dfrac{4a-5}{7}$ неотрицательна. + В ответ запишите наименьшее натуральное значение переменной.`, + sol: `Метод решения линейного неравенства с дробями: приводим дроби к общему знаменателю, затем избавляемся от знаменателя (если он положителен — знак сохраняется). +
Шаг 1. По условию разность дробей неотрицательна, значит составим неравенство: +$$\\dfrac{4a-3}{5} - \\dfrac{4a-5}{7} \\geq 0$$ +Шаг 2. Приводим к общему знаменателю $35$: +$$\\dfrac{7(4a-3) - 5(4a-5)}{35} \\geq 0$$ +Шаг 3. Раскрываем скобки в числителе: +$$7(4a-3) = 28a - 21, \\quad 5(4a-5) = 20a - 25$$ +$$28a - 21 - (20a - 25) = 28a - 21 - 20a + 25 = 8a + 4$$ +Неравенство принимает вид: +$$\\dfrac{8a + 4}{35} \\geq 0$$ +Шаг 4. Так как $35 \\gt 0$, можно умножить обе части на $35$ без изменения знака: +$$8a + 4 \\geq 0 \\implies 8a \\geq -4 \\implies a \\geq -\\dfrac{1}{2}$$ +Шаг 5. Среди натуральных чисел ($1, 2, 3,\\ldots$) условию $a \\geq -\\dfrac{1}{2}$ удовлетворяют все. Наименьшее натуральное число — это $a = 1$. +
Ответ: $1$
` + }, + { + text: `Найдите значение выражения + $\\left(\\dfrac{4}{3-\\sqrt{5}}\\right)^{\\!2} - \\left(\\dfrac{6-5\\sqrt{6}}{5-\\sqrt{6}}\\right)^{\\!2}$. + В ответ запишите число, противоположное найденному.`, + sol: `Метод рационализации знаменателя: чтобы избавиться от иррациональности в знаменателе вида $a - \\sqrt{b}$, умножаем числитель и знаменатель на сопряжённое $a + \\sqrt{b}$ и применяем формулу разности квадратов: $(a-\\sqrt{b})(a+\\sqrt{b}) = a^2 - b$. +
Шаг 1. Упростим первую дробь $\\dfrac{4}{3 - \\sqrt{5}}$, умножив числитель и знаменатель на сопряжённое $3 + \\sqrt{5}$: +$$\\dfrac{4}{3-\\sqrt{5}} = \\dfrac{4(3+\\sqrt{5})}{(3-\\sqrt{5})(3+\\sqrt{5})} = \\dfrac{4(3+\\sqrt{5})}{9 - 5} = \\dfrac{4(3+\\sqrt{5})}{4} = 3 + \\sqrt{5}$$ +Шаг 2. Упростим вторую дробь $\\dfrac{6 - 5\\sqrt{6}}{5 - \\sqrt{6}}$, умножив на сопряжённое $5 + \\sqrt{6}$: +$$\\dfrac{(6-5\\sqrt{6})(5+\\sqrt{6})}{(5-\\sqrt{6})(5+\\sqrt{6})} = \\dfrac{30 + 6\\sqrt{6} - 25\\sqrt{6} - 5\\cdot 6}{25 - 6}$$ +В числителе: $30 - 30 + (6 - 25)\\sqrt{6} = -19\\sqrt{6}$. В знаменателе: $19$. Значит: +$$\\dfrac{-19\\sqrt{6}}{19} = -\\sqrt{6}$$ +Шаг 3. Подставим в исходное выражение, применяя формулу квадрата суммы $(a+b)^2 = a^2 + 2ab + b^2$: +$$(3+\\sqrt{5})^2 - (-\\sqrt{6})^2 = (9 + 6\\sqrt{5} + 5) - 6 = 14 + 6\\sqrt{5} - 6 = 8 + 6\\sqrt{5}$$ +Шаг 4. По условию надо записать число, противоположное найденному. Противоположное к $8 + 6\\sqrt{5}$ — это $-(8 + 6\\sqrt{5})$. +
Ответ: $-(8+6\\sqrt{5})$
` + }, + { + text: `В ботаническом саду ландшафтный дизайнер решил разместить кусты роз так, + чтобы в каждом ряду было одинаковое количество кустов, + при этом рядов — на $8$ больше, чем кустов в каждом ряду. + Определите, можно ли на клумбе посадить $128$ кустов роз. Ответ обоснуйте.`, + sol: `Метод введения переменной и составления квадратного уравнения. +
Шаг 1. Пусть $x$ — количество кустов в каждом ряду ($x$ — натуральное число). По условию рядов на $8$ больше, значит число рядов равно $x + 8$. +
Шаг 2. Общее число кустов = (число в одном ряду) $\\times$ (число рядов). По условию оно равно $128$, поэтому: +$$x(x + 8) = 128 \\implies x^2 + 8x - 128 = 0$$ +Шаг 3. Решаем квадратное уравнение по формуле дискриминанта $D = b^2 - 4ac$: +$$D = 8^2 - 4\\cdot 1\\cdot(-128) = 64 + 512 = 576 = 24^2$$ +$$x = \\dfrac{-8 \\pm 24}{2} \\implies x_1 = 8, \\quad x_2 = -16$$ +Шаг 4. Так как $x$ — количество кустов, оно должно быть натуральным, поэтому $x_2 = -16$ не подходит. Остаётся $x = 8$. +
Шаг 5. Проверка: $8$ кустов в каждом ряду, рядов $8 + 8 = 16$, всего кустов $8\\cdot 16 = 128$ $\\checkmark$. +
Ответ: да, можно — $8$ кустов в ряду и $16$ рядов
` + }, + { + text: `Гипотенуза прямоугольного треугольника равна $10$ см, радиус вписанной окружности — $2$ см. + Найдите площадь треугольника.`, + sol: `Формула радиуса вписанной окружности прямоугольного треугольника: с катетами $a$, $b$ и гипотенузой $c$: +$$r = \\dfrac{a + b - c}{2}$$ +Формула площади через вписанную окружность: $S = r\\cdot s$, где $s = \\dfrac{a + b + c}{2}$ — полупериметр. +
Шаг 1. Из формулы радиуса находим сумму катетов: +$$2 = \\dfrac{a + b - 10}{2} \\implies a + b - 10 = 4 \\implies a + b = 14$$ + + + + + + + + C + A + B + I + $b$ + $a$ + r=2 + +Шаг 2. Считаем полупериметр: +$$s = \\dfrac{a + b + c}{2} = \\dfrac{14 + 10}{2} = 12\\text{ см}$$ +Шаг 3. Находим площадь: +$$S = r\\cdot s = 2\\cdot 12 = 24\\text{ см}^2$$ +Проверка: по теореме Пифагора $a^2 + b^2 = c^2 = 100$. Из $(a + b)^2 = 14^2 = 196$ получаем $a^2 + 2ab + b^2 = 196$, значит $2ab = 196 - 100 = 96$, то есть $ab = 48$. Площадь прямоугольного треугольника $= \\dfrac{ab}{2} = 24$ ✓ +
Ответ: $24$ см²
` + }, + { + text: `Решите уравнение $x(x-1)(x-3)(x-4) = 40$. + В ответ запишите корни уравнения, удовлетворяющие неравенству $|x| < 5$.`, + sol: `Идея решения: в уравнении вида $(x-a)(x-b)(x-c)(x-d) = k$ удобно группировать множители так, чтобы суммы корней внутри пар были одинаковы. +
Шаг 1. Сгруппируем так: $\\{0,\\,4\\}$ и $\\{1,\\,3\\}$ — обе пары имеют сумму $4$. +$$[x(x - 4)]\\cdot[(x - 1)(x - 3)] = 40$$ +Раскрываем скобки в каждой паре: +$$x(x - 4) = x^2 - 4x$$ +$$(x - 1)(x - 3) = x^2 - 4x + 3$$ +Тогда уравнение принимает вид: +$$(x^2 - 4x)(x^2 - 4x + 3) = 40$$ +Шаг 2. Замена переменной. Пусть $t = x^2 - 4x$: +$$t(t + 3) = 40 \\implies t^2 + 3t - 40 = 0$$ +Шаг 3. Решаем по формуле дискриминанта: +$$D = 3^2 - 4\\cdot 1\\cdot(-40) = 9 + 160 = 169 = 13^2$$ +$$t = \\dfrac{-3 \\pm 13}{2} \\implies t_1 = 5, \\quad t_2 = -8$$ +Шаг 4. Случай $t = 5$: $x^2 - 4x = 5 \\implies x^2 - 4x - 5 = 0$. +
По теореме Виета (сумма $4$, произведение $-5$): корни $5$ и $-1$. +$$(x - 5)(x + 1) = 0 \\implies x = 5\\text{ или } x = -1$$ +Шаг 5. Случай $t = -8$: $x^2 - 4x = -8 \\implies x^2 - 4x + 8 = 0$. +$$D = 16 - 32 = -16 \\lt 0$$ +Дискриминант отрицателен, поэтому вещественных корней нет. +
Шаг 6. Проверка условия $|x| \\lt 5$: +
    +
  • $x = 5$: $|5| = 5$, неравенство $|x| \\lt 5$ строгое, поэтому не подходит.
    • +
  • $x = -1$: $|-1| = 1 \\lt 5$ ✓
    • +
+
Ответ: $x = -1$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v42.js b/frontend/js/exam9/variants/v42.js new file mode 100644 index 0000000..435f148 --- /dev/null +++ b/frontend/js/exam9/variants/v42.js @@ -0,0 +1,190 @@ +VARIANTS[42] = { + label: "Вариант 42", + tasks: [ + { + text: `Определите верное равенство:`, + opts: [ + ["а", "$9\\% = 90$"], ["б", "$9\\% = 9$"], ["в", "$9\\% = 0{,}9$"], + ["г", "$9\\% = 0{,}09$"], ["д", "$9\\% = 0{,}009$"], + ], + sol: `$9\\% = \\dfrac{9}{100} = 0{,}09$ +
Ответ: г) $9\\%=0{,}09$
` + }, + { + text: `Запишите одночлен $2m^5n \\cdot \\left(-\\dfrac{1}{2}m^4\\right)$ в стандартном виде:`, + opts: [ + ["а", "$-mn$"], ["б", "$-m^9n$"], ["в", "$-mn^9$"], + ["г", "$\\dfrac{1}{2}mn$"], ["д", "$4m^9n$"], + ], + sol: `$$2m^5n\\cdot\\left(-\\dfrac{1}{2}m^4\\right) = \\left(2\\cdot\\left(-\\dfrac{1}{2}\\right)\\right)\\cdot m^{5+4}\\cdot n = -m^9n$$ +
Ответ: б) $-m^9n$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "радиус вписанной окружности прямоугольного треугольника с катетами $a$, $b$ и гипотенузой $c$ находится по формуле $r = \\dfrac{a+b-c}{2}$;"], + ["б", "площадь прямоугольника со сторонами $a$, $b$ находится по формуле $S = ab$;"], + ["в", "из данной точки, не лежащей на данной прямой, к данной прямой можно провести только один перпендикуляр;"], + ["г", "если в треугольнике $ABC$ сторона $AC$ — наибольшая, то угол $A$ — наибольший?"], + ], + sol: `
    +
  • а) $r=\\dfrac{a+b-c}{2}$ — верно
    • +
  • б) $S=ab$ для прямоугольника — верно
    • +
  • в) один перпендикуляр — верно
    • +
  • г) $AC$ напротив $\\angle B$ — наибольший $\\angle B$, не $\\angle A$ — НЕВЕРНО
    • +
+
Ответ: г)
` + }, + { + text: `График функции $y = kx + 5$ проходит через точку $K\\!\\left(-\\dfrac{1}{5};\\; 25\\right)$. + Найдите коэффициент $k$.`, + sol: `Подставляем $K\\!\\left(-\\dfrac{1}{5};\\;25\\right)$: $25=k\\cdot\\left(-\\dfrac{1}{5}\\right)+5 \\implies 20=-\\dfrac{k}{5} \\implies k=-100$. +
Ответ: $k = -100$
` + }, + { + text: `Диагонали параллелограмма $ABCD$ пересекаются в точке $O$, + площадь треугольника $COD$ равна $12$ см². + Высота, проведённая из вершины $C$ к $AB$, равна $8$ см. + Найдите длину стороны $CD$ параллелограмма.`, + sol: `Свойство диагоналей параллелограмма: диагонали параллелограмма делят его на $4$ треугольника равной площади (равновеликие треугольники). +
Шаг 1. По этому свойству: +$$S_{ABCD} = 4\\cdot S_{COD} = 4\\cdot 12 = 48\\text{ см}^2$$ + + + + + + + + + + + A + B + C + D + O + 12 + h=8 + +Шаг 2. Формула площади параллелограмма: $S = a\\cdot h$, где $a$ — сторона, $h$ — высота, проведённая к этой стороне. +
Высота, проведённая из $C$ к $AB$, и есть высота параллелограмма к стороне $AB$: +$$S_{ABCD} = AB\\cdot h \\implies 48 = AB\\cdot 8 \\implies AB = 6\\text{ см}$$ +Шаг 3. По свойству параллелограмма противоположные стороны равны: +$$CD = AB = 6\\text{ см}$$ +
Ответ: $CD = 6$ см
` + }, + { + text: `Определите, при каких значениях переменной разность дробей + $\\dfrac{2b-5}{4}$ и $\\dfrac{4b-3}{6}$ неположительна. + В ответ запишите наименьшее целое значение переменной.`, + sol: `Метод решения линейного неравенства с дробями: приводим дроби к общему знаменателю, затем избавляемся от знаменателя (если он положителен — знак сохраняется). +
Шаг 1. По условию разность дробей неположительна: +$$\\dfrac{2b-5}{4} - \\dfrac{4b-3}{6} \\leq 0$$ +Шаг 2. Приводим к общему знаменателю $12$: +$$\\dfrac{3(2b-5) - 2(4b-3)}{12} \\leq 0$$ +Шаг 3. Раскрываем скобки: +$$3(2b - 5) = 6b - 15, \\quad 2(4b - 3) = 8b - 6$$ +$$6b - 15 - (8b - 6) = 6b - 15 - 8b + 6 = -2b - 9$$ +Неравенство принимает вид: +$$\\dfrac{-2b - 9}{12} \\leq 0$$ +Шаг 4. Так как $12 \\gt 0$, умножим обе части на $12$ без смены знака: +$$-2b - 9 \\leq 0 \\implies -2b \\leq 9$$ +Делим на $-2$ (отрицательное число — знак неравенства меняется): +$$b \\geq -\\dfrac{9}{2} = -4{,}5$$ +Шаг 5. Среди целых чисел условию $b \\geq -4{,}5$ удовлетворяют $-4, -3, -2, \\ldots$. Наименьшее целое — это $b = -4$. +
Ответ: $-4$
` + }, + { + text: `Найдите значение выражения + $\\left(\\dfrac{7}{3-\\sqrt{2}}\\right)^{\\!2} - \\left(\\dfrac{3-4\\sqrt{3}}{4-\\sqrt{3}}\\right)^{\\!2}$. + В ответ запишите число, противоположное найденному.`, + sol: `Метод рационализации знаменателя: чтобы избавиться от иррациональности в знаменателе вида $a - \\sqrt{b}$, умножаем на сопряжённое $a + \\sqrt{b}$ и применяем формулу разности квадратов: $(a-\\sqrt{b})(a+\\sqrt{b}) = a^2 - b$. +
Шаг 1. Упростим первую дробь $\\dfrac{7}{3 - \\sqrt{2}}$: +$$\\dfrac{7}{3-\\sqrt{2}} = \\dfrac{7(3+\\sqrt{2})}{(3-\\sqrt{2})(3+\\sqrt{2})} = \\dfrac{7(3+\\sqrt{2})}{9 - 2} = \\dfrac{7(3+\\sqrt{2})}{7} = 3 + \\sqrt{2}$$ +Шаг 2. Упростим вторую дробь $\\dfrac{3 - 4\\sqrt{3}}{4 - \\sqrt{3}}$, умножив числитель и знаменатель на сопряжённое $4 + \\sqrt{3}$: +$$\\dfrac{(3 - 4\\sqrt{3})(4 + \\sqrt{3})}{(4 - \\sqrt{3})(4 + \\sqrt{3})} = \\dfrac{12 + 3\\sqrt{3} - 16\\sqrt{3} - 4\\cdot 3}{16 - 3}$$ +В числителе: $12 - 12 + (3 - 16)\\sqrt{3} = -13\\sqrt{3}$. В знаменателе: $13$. Получаем: +$$\\dfrac{-13\\sqrt{3}}{13} = -\\sqrt{3}$$ +Шаг 3. Подставим в исходное выражение, используя формулу квадрата суммы $(a+b)^2 = a^2 + 2ab + b^2$: +$$(3 + \\sqrt{2})^2 - (-\\sqrt{3})^2 = (9 + 6\\sqrt{2} + 2) - 3 = 11 + 6\\sqrt{2} - 3 = 8 + 6\\sqrt{2}$$ +Шаг 4. По условию надо записать число, противоположное найденному. Противоположное к $8 + 6\\sqrt{2}$ — это $-(8 + 6\\sqrt{2})$. +
Ответ: $-(8+6\\sqrt{2})$
` + }, + { + text: `В ботаническом саду ландшафтный дизайнер решил разместить кусты пионов так, + чтобы в каждом ряду было одинаковое количество кустов, + при этом рядов — на $7$ меньше, чем кустов в каждом ряду. + Определите, можно ли на клумбе посадить $60$ кустов пионов. Ответ обоснуйте.`, + sol: `Метод введения переменной и составления квадратного уравнения. +
Шаг 1. Пусть $x$ — количество кустов в каждом ряду ($x$ — натуральное число). По условию рядов на $7$ меньше, значит число рядов равно $x - 7$ (это должно быть положительно, то есть $x \\gt 7$). +
Шаг 2. Общее число кустов = (число в одном ряду) $\\times$ (число рядов). По условию оно равно $60$, значит: +$$x(x - 7) = 60 \\implies x^2 - 7x - 60 = 0$$ +Шаг 3. Решаем по формуле дискриминанта $D = b^2 - 4ac$: +$$D = (-7)^2 - 4\\cdot 1\\cdot(-60) = 49 + 240 = 289 = 17^2$$ +$$x = \\dfrac{7 \\pm 17}{2} \\implies x_1 = 12, \\quad x_2 = -5$$ +Шаг 4. Корень $x_2 = -5$ не подходит, так как $x$ должно быть натуральным. Остаётся $x = 12$. +
Шаг 5. Проверка: $12$ кустов в каждом ряду, рядов $12 - 7 = 5$, всего кустов $12\\cdot 5 = 60$ $\\checkmark$. +
Ответ: да, можно — $12$ кустов в ряду и $5$ рядов
` + }, + { + text: `Гипотенуза прямоугольного треугольника равна $13$ см, радиус вписанной окружности — $2$ см. + Найдите площадь треугольника.`, + sol: `Формула радиуса вписанной окружности прямоугольного треугольника: с катетами $a$, $b$ и гипотенузой $c$: +$$r = \\dfrac{a + b - c}{2}$$ +Формула площади через вписанную окружность: $S = r\\cdot s$, где $s = \\dfrac{a + b + c}{2}$ — полупериметр. +
Шаг 1. Из формулы радиуса находим сумму катетов: +$$2 = \\dfrac{a + b - 13}{2} \\implies a + b - 13 = 4 \\implies a + b = 17$$ + + + + + + + + C + A + B + I + $b$ + $a$ + r=2 + +Шаг 2. Считаем полупериметр: +$$s = \\dfrac{a + b + c}{2} = \\dfrac{17 + 13}{2} = 15\\text{ см}$$ +Шаг 3. Находим площадь: +$$S = r\\cdot s = 2\\cdot 15 = 30\\text{ см}^2$$ +Проверка: по теореме Пифагора $a^2 + b^2 = 169$. Из $(a + b)^2 = 17^2 = 289$ получаем $a^2 + 2ab + b^2 = 289$, значит $2ab = 289 - 169 = 120$, то есть $ab = 60$. Площадь прямоугольного треугольника $= \\dfrac{ab}{2} = 30$ ✓ +
Ответ: $30$ см²
` + }, + { + text: `Решите уравнение $x(x-1)(x-2)(x-3) = 24$. + В ответ запишите корни уравнения, удовлетворяющие неравенству $|x| < 4$.`, + sol: `Идея решения: в уравнении вида $(x-a)(x-b)(x-c)(x-d) = k$ группируем множители так, чтобы суммы корней внутри пар были одинаковы. +
Шаг 1. Сгруппируем: $\\{0,\\,3\\}$ и $\\{1,\\,2\\}$ — обе пары имеют сумму $3$. +$$[x(x - 3)]\\cdot[(x - 1)(x - 2)] = 24$$ +Раскрываем скобки в каждой паре: +$$x(x - 3) = x^2 - 3x$$ +$$(x - 1)(x - 2) = x^2 - 3x + 2$$ +Уравнение принимает вид: +$$(x^2 - 3x)(x^2 - 3x + 2) = 24$$ +Шаг 2. Замена переменной. Пусть $t = x^2 - 3x$: +$$t(t + 2) = 24 \\implies t^2 + 2t - 24 = 0$$ +Шаг 3. Решаем по формуле дискриминанта: +$$D = 2^2 - 4\\cdot 1\\cdot(-24) = 4 + 96 = 100 = 10^2$$ +$$t = \\dfrac{-2 \\pm 10}{2} \\implies t_1 = 4, \\quad t_2 = -6$$ +Шаг 4. Случай $t = 4$: $x^2 - 3x = 4 \\implies x^2 - 3x - 4 = 0$. +
По теореме Виета (сумма $3$, произведение $-4$): корни $4$ и $-1$. +$$(x - 4)(x + 1) = 0 \\implies x = 4 \\text{ или } x = -1$$ +Шаг 5. Случай $t = -6$: $x^2 - 3x = -6 \\implies x^2 - 3x + 6 = 0$. +$$D = 9 - 24 = -15 \\lt 0$$ +Дискриминант отрицателен — вещественных корней нет. +
Шаг 6. Проверка условия $|x| \\lt 4$: +
    +
  • $x = 4$: $|4| = 4$, строгое неравенство $|x| \\lt 4$ не выполняется.
    • +
  • $x = -1$: $|-1| = 1 \\lt 4$ ✓
    • +
+
Ответ: $x = -1$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v43.js b/frontend/js/exam9/variants/v43.js new file mode 100644 index 0000000..aed03ce --- /dev/null +++ b/frontend/js/exam9/variants/v43.js @@ -0,0 +1,244 @@ +VARIANTS[43] = { + label: "Вариант 43", + tasks: [ + { + text: `Выберите функцию, график которой изображён на рисунке:`, + figure: ` + + + + + + + + + + + + + + + + + + + + + + x + y + 0 + 2 + 4 + 2 + 4 + 6 + + + + + + (2; 2) +`, + opts: [ + ["а", "$y = (x+2)^2 - 2$"], ["б", "$y = (x+2)^2 + 2$"], ["в", "$y = (x-2)^2 + 3$"], + ["г", "$y = (x-2)^2 + 2$"], ["д", "$y = (x-2)^2 - 2$"], + ], + sol: `Свойство параболы $y=(x-a)^2+b$: вершина находится в точке $(a;\\,b)$, ветви направлены вверх (коэффициент при $x^2$ положительный). +
Шаг 1. По рисунку определяем координаты вершины: $(2;\\,2)$. Значит $a=2$, $b=2$. +
Шаг 2. Подставляем в общую формулу: $y=(x-2)^2+2$. +
Проверка: при $x=0$ получаем $y=4+2=6$ — точка $(0;\\,6)$ на графике; при $x=4$ получаем $y=4+2=6$ — точка $(4;\\,6)$. Парабола симметрична относительно прямой $x=2$ — это и есть ось симметрии через вершину. +
Ответ: г) $y=(x-2)^2+2$
` + }, + { + text: `Результат сокращения дроби $\\dfrac{18ab - 9a}{9ab}$ равен:`, + opts: [ + ["а", "$\\dfrac{2ab-a}{ab}$"], ["б", "$\\dfrac{2b-1}{b}$"], ["в", "$2b-1$"], + ["г", "$\\dfrac{b-1}{b}$"], ["д", "$2ab - 9a$"], + ], + sol: `Выносим $9a$ за скобку в числителе: +$$\\dfrac{18ab-9a}{9ab} = \\dfrac{9a(2b-1)}{9ab} = \\dfrac{2b-1}{b}\\quad(a\\neq0,\\,b\\neq0)$$ +
Ответ: б) $\\dfrac{2b-1}{b}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "в любой треугольник можно вписать окружность;"], + ["б", "в равнобедренном треугольнике углы при основании равны;"], + ["в", "если у четырёхугольника все углы прямые, то это прямоугольник;"], + ["г", "прямой угол равен $100^{\\circ}$?"], + ], + sol: `
    +
  • а) В любой треугольник вписывается окружность — верно
    • +
  • б) Равнобедренный: углы при основании равны — верно
    • +
  • в) Все углы прямые ⟹ прямоугольник — верно
    • +
  • г) «Прямой угол $=100°$» — НЕВЕРНО. Прямой угол $= 90°$.
    • +
+
Ответ: г)
` + }, + { + text: `Найдите количество целых решений системы неравенств + $$\\begin{cases} x < -10, \\\\[4pt] x > -15. \\end{cases}$$`, + sol: `Система: $-15 < x < -10$. Целые числа в этом промежутке: +$$-14,\\;-13,\\;-12,\\;-11$$ +
Ответ: $4$ целых решения
` + }, + { + text: `Найдите площадь треугольника со сторонами $5$ см, $5$ см и $6$ см.`, + sol: `Свойство равнобедренного треугольника: высота, проведённая к основанию, является также медианой (т.е. делит основание пополам). +
Формула площади треугольника: $S = \\dfrac{1}{2}\\cdot a\\cdot h$, где $a$ — основание, $h$ — высота к нему. +
Шаг 1. У нашего треугольника две стороны по $5$ см и одна $6$ см — он равнобедренный с основанием $6$ см. Высота $CM$, опущенная на основание, делит его пополам: $AM = MB = 3$ см. + + + + + + A + B + C + M + 5 + 5 + 6 + h + +Шаг 2. Найдём высоту $h = CM$ по теореме Пифагора в прямоугольном треугольнике $ACM$: +$$AC^2 = AM^2 + CM^2 \\implies 5^2 = 3^2 + h^2$$ +$$h^2 = 25 - 9 = 16 \\implies h = \\sqrt{16} = 4\\text{ см}$$ +Шаг 3. Считаем площадь: +$$S = \\dfrac{1}{2}\\cdot 6\\cdot 4 = 12\\text{ см}^2$$ +

Альтернативный способ — формула Герона. +
Формула Герона: $S = \\sqrt{p(p-a)(p-b)(p-c)}$, где $p = \\dfrac{a+b+c}{2}$ — полупериметр треугольника со сторонами $a$, $b$, $c$. +
Шаг 1. Полупериметр: $p = \\dfrac{5+5+6}{2} = \\dfrac{16}{2} = 8$ см. +
Шаг 2. Подставляем в формулу: +$$S = \\sqrt{8\\cdot(8-5)\\cdot(8-5)\\cdot(8-6)} = \\sqrt{8\\cdot 3\\cdot 3\\cdot 2} = \\sqrt{144} = 12\\text{ см}^2$$ +Оба способа дают один и тот же ответ. +
Ответ: $12$ см²
` + }, + { + text: `Упростите выражение $\\sqrt{7}\\cdot(\\sqrt{64} + \\sqrt{112} - 5\\sqrt{7})\\cdot(-\\sqrt{28})$.`, + sol: `Свойство квадратного корня: $\\sqrt{a\\cdot b} = \\sqrt{a}\\cdot\\sqrt{b}$ (для $a,b\\geq 0$), а также $\\sqrt{a}\\cdot\\sqrt{a} = a$ (для $a\\geq 0$). +
Шаг 1. Упрощаем каждый корень в скобках, вынося полный квадрат из-под корня: +$$\\sqrt{64} = 8$$ +$$\\sqrt{112} = \\sqrt{16\\cdot 7} = \\sqrt{16}\\cdot\\sqrt{7} = 4\\sqrt{7}$$ +$$\\sqrt{28} = \\sqrt{4\\cdot 7} = \\sqrt{4}\\cdot\\sqrt{7} = 2\\sqrt{7}$$ +Шаг 2. Подставляем и упрощаем выражение в скобках, приводя подобные слагаемые $4\\sqrt{7}$ и $-5\\sqrt{7}$: +$$8 + 4\\sqrt{7} - 5\\sqrt{7} = 8 - \\sqrt{7}$$ +Исходное выражение принимает вид: +$$\\sqrt{7}\\cdot(8 - \\sqrt{7})\\cdot(-2\\sqrt{7})$$ +Шаг 3. Перемножим крайние множители $\\sqrt{7}$ и $-2\\sqrt{7}$: +$$\\sqrt{7}\\cdot(-2\\sqrt{7}) = -2\\cdot(\\sqrt{7})^2 = -2\\cdot 7 = -14$$ +Шаг 4. Умножаем результат на оставшуюся скобку, раскрывая её: +$$-14\\cdot(8 - \\sqrt{7}) = -14\\cdot 8 + 14\\sqrt{7} = -112 + 14\\sqrt{7} = 14(\\sqrt{7} - 8)$$ +
Ответ: $14(\\sqrt{7}-8)$
` + }, + { + text: `Найдите область определения функции + $y = \\sqrt{12 - 6x} - \\dfrac{3}{x^2 - 4}$.`, + sol: `Правила нахождения области определения: +
1) Подкоренное выражение чётной степени должно быть неотрицательным: $\\sqrt{f(x)}$ определён при $f(x) \\geq 0$. +
2) Знаменатель дроби не может равняться нулю: $\\dfrac{1}{g(x)}$ определена при $g(x) \\neq 0$. +
В нашей функции присутствуют оба элемента — выписываем оба условия. +
Шаг 1. Подкоренное выражение $12 - 6x$ должно быть $\\geq 0$: +$$12 - 6x \\geq 0 \\implies 6x \\leq 12 \\implies x \\leq 2$$ +Шаг 2. Знаменатель $x^2 - 4$ должен быть $\\neq 0$: +$$x^2 - 4 \\neq 0 \\implies x^2 \\neq 4 \\implies x \\neq \\pm 2$$ +Шаг 3. Объединяем условия: $x \\leq 2$ и $x \\neq -2$ и $x \\neq 2$. +
Условие $x \\leq 2$ с исключением $x = 2$ превращается в $x \\lt 2$. С учётом $x \\neq -2$: +$$x \\in (-\\infty;\\,-2)\\cup(-2;\\,2)$$ +
Ответ: $(-\\infty;\\,-2)\\cup(-2;\\,2)$
` + }, + { + text: `Найдите значение выражения $81x_0$, где $x_0$ — наибольший корень уравнения + $$\\dfrac{x^2}{4x^2+4x+1} - \\dfrac{6x}{2x+1} + 5 = 0.$$`, + sol: `Формула квадрата суммы: $(a + b)^2 = a^2 + 2ab + b^2$. +
Шаг 1. Замечаем, что $4x^2 + 4x + 1 = (2x)^2 + 2\\cdot 2x\\cdot 1 + 1^2 = (2x + 1)^2$. +
Уравнение принимает вид: +$$\\dfrac{x^2}{(2x + 1)^2} - \\dfrac{6x}{2x + 1} + 5 = 0$$ +ОДЗ: $2x + 1 \\neq 0$, то есть $x \\neq -\\dfrac{1}{2}$. +
Шаг 2. Замена переменной. Заметим, что $\\dfrac{x^2}{(2x + 1)^2} = \\left(\\dfrac{x}{2x + 1}\\right)^2$. +
Пусть $t = \\dfrac{x}{2x + 1}$. Уравнение становится: +$$t^2 - 6t + 5 = 0$$ +Шаг 3. По теореме Виета (сумма $6$, произведение $5$): корни $1$ и $5$. +$$(t - 1)(t - 5) = 0 \\implies t = 1 \\text{ или } t = 5$$ +Шаг 4. Возвращаемся к $x$. +
— При $t = 1$: $\\dfrac{x}{2x + 1} = 1 \\implies x = 2x + 1 \\implies -x = 1 \\implies x = -1$. +
— При $t = 5$: $\\dfrac{x}{2x + 1} = 5 \\implies x = 5(2x + 1) = 10x + 5 \\implies -9x = 5 \\implies x = -\\dfrac{5}{9}$. +
Шаг 5. Проверка ОДЗ. Оба значения $\\neq -\\dfrac{1}{2}$ ✓. +
Шаг 6. Сравним корни: $-\\dfrac{5}{9} \\approx -0{,}56$, а $-1$ меньше. Наибольший корень: +$$x_0 = -\\dfrac{5}{9}$$ +$$81x_0 = 81\\cdot\\left(-\\dfrac{5}{9}\\right) = -\\dfrac{81\\cdot 5}{9} = -9\\cdot 5 = -45$$ +
Ответ: $-45$
` + }, + { + text: `Найдите площадь описанной равнобедренной трапеции, если точка касания вписанной + в неё окружности делит боковую сторону на отрезки, равные $4$ см и $9$ см.`, + sol: `Шаг 1. Основания трапеции. +
Боковая сторона $= 4+9 = 13$ см. По свойству касательных из одной точки: от каждой вершины оба касательных отрезка равны. +
Обозначим: от вершин большего основания — по $9$, от вершин меньшего — по $4$. + + + + + + + + + + + + + + + + + + + + + + + A + D + B + C + O + + 9 + 4 + 9 + 4 + + AD = 18 + BC = 8 + + r = 6 + +$$AD = 9+9 = 18\\text{ см}, \\quad BC = 4+4 = 8\\text{ см}$$ +Шаг 2. Высота трапеции. +
Горизонтальный выступ ноги: $\\dfrac{AD-BC}{2} = \\dfrac{18-8}{2} = 5$ см. +$$h = \\sqrt{13^2-5^2} = \\sqrt{169-25} = \\sqrt{144} = 12\\text{ см}$$ +Шаг 3. Площадь. +$$S = \\dfrac{AD+BC}{2}\\cdot h = \\dfrac{18+8}{2}\\cdot12 = 13\\cdot12 = 156\\text{ см}^2$$ +
Ответ: $156$ см²
` + }, + { + text: `В компанию поступил заказ на укладку $240$ м² напольной плитки. + Плиточник принял решение укладывать на $10$ м² в день больше, чем запланировал ранее. + В результате работа была закончена на $4$ дня раньше установленного срока. + Успеет ли плиточник выполнить заказ за $10$ рабочих дней, если будет работать + по первоначальному плану? Ответ обоснуйте.`, + sol: `Пусть плановая выработка $= x$ м²/день. +
По плану: $\\dfrac{240}{x}$ дней. С ускорением: $\\dfrac{240}{x+10}$ дней, на $4$ меньше. +$$\\dfrac{240}{x+10} = \\dfrac{240}{x} - 4$$ +Умножаем на $x(x+10)$: +$$240x = 240(x+10) - 4x(x+10)$$ +$$0 = 2400 - 4x^2 - 40x \\implies x^2+10x-600=0$$ +$$D = 100+2400 = 2500 = 50^2 \\implies x = \\dfrac{-10+50}{2} = 20\\text{ м²/день}$$ +Плановый срок: $\\dfrac{240}{20} = 12$ дней. +
Проверка: при $30$ м²/день: $240\\div30=8$ дней, $12-8=4$ ✓ +
Ответ на вопрос: за $10$ дней при выработке $20$ м²/день плиточник уложит $10\\times20=200$ м² $<240$ м². +
Ответ: нет, не успеет — за $10$ дней уложит лишь $200$ м² из $240$ м²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v44.js b/frontend/js/exam9/variants/v44.js new file mode 100644 index 0000000..b89883b --- /dev/null +++ b/frontend/js/exam9/variants/v44.js @@ -0,0 +1,242 @@ +VARIANTS[44] = { + label: "Вариант 44", + tasks: [ + { + text: `Выберите функцию, график которой изображён на рисунке:`, + figure: ` + + + + + + + + + + + + + + + + + + + + x + y + 0 + 4 + 8 + 2 + -2 + -4 + + + + + + (4; -2) +`, + opts: [ + ["а", "$y = (x+2)^2 - 2$"], ["б", "$y = (x+2)^2 + 2$"], ["в", "$y = (x-4)^2 - 2$"], + ["г", "$y = (x-4)^2 + 2$"], ["д", "$y = (x+4)^2 + 2$"], + ], + sol: `Свойство параболы $y=(x-a)^2+b$: вершина находится в точке $(a;\\,b)$, ветви направлены вверх (коэффициент при $x^2$ положительный). +
Шаг 1. По рисунку определяем координаты вершины: $(4;\\,-2)$. Значит $a=4$, $b=-2$. +
Шаг 2. Подставляем в общую формулу: $y=(x-4)^2+(-2)=(x-4)^2-2$. +
Проверка: при $x=2$ получаем $y=4-2=2$ — точка $(2;\\,2)$ на графике; при $x=6$ получаем $y=4-2=2$ — точка $(6;\\,2)$. Парабола симметрична относительно прямой $x=4$ — это ось симметрии через вершину. +
Ответ: в) $y=(x-4)^2-2$
` + }, + { + text: `Результат сокращения дроби $\\dfrac{7mn - 35m}{7mn}$ равен:`, + opts: [ + ["а", "$\\dfrac{mn-5m}{mn}$"], ["б", "$\\dfrac{n-5}{n}$"], ["в", "$n-5$"], + ["г", "$\\dfrac{n-4}{n}$"], ["д", "$1 - 35m$"], + ], + sol: `Выносим $7m$ за скобку в числителе: +$$\\dfrac{7mn-35m}{7mn} = \\dfrac{7m(n-5)}{7mn} = \\dfrac{n-5}{n}\\quad(m\\neq0,\\,n\\neq0)$$ +
Ответ: б) $\\dfrac{n-5}{n}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "около любого треугольника можно описать окружность;"], + ["б", "если в треугольнике равны два угла, то треугольник равнобедренный;"], + ["в", "если у четырёхугольника два угла прямые, то это всегда прямоугольник;"], + ["г", "развёрнутый угол равен $180^{\\circ}$?"], + ], + sol: `
    +
  • а) Около любого треугольника можно описать окружность — верно
    • +
  • б) Два равных угла ⟹ равнобедренный треугольник — верно
    • +
  • в) Четырёхугольник с двумя прямыми углами — это необязательно прямоугольник. Например, прямоугольная трапеция имеет два прямых угла, но не является прямоугольником — НЕВЕРНО
    • +
  • г) Развёрнутый угол $=180°$ — верно
    • +
+
Ответ: в)
` + }, + { + text: `Найдите количество целых решений системы неравенств + $$\\begin{cases} x < 10, \\\\[4pt] x > 3. \\end{cases}$$`, + sol: `Система: $3 < x < 10$. Целые числа в этом промежутке: +$$4,\\;5,\\;6,\\;7,\\;8,\\;9$$ +
Ответ: $6$ целых решений
` + }, + { + text: `Найдите площадь треугольника со сторонами $13$ см, $13$ см и $10$ см.`, + sol: `Свойство равнобедренного треугольника: высота, проведённая к основанию, является также медианой (делит основание пополам). +
Формула площади треугольника: $S = \\dfrac{1}{2}\\cdot a\\cdot h$, где $a$ — основание, $h$ — высота к нему. +
Шаг 1. У нашего треугольника две стороны по $13$ см и одна $10$ см — он равнобедренный с основанием $10$ см. Высота $CM$, опущенная на основание, делит его пополам: $AM = MB = 5$ см. + + + + + + A + B + C + M + 13 + 13 + 10 + h + +Шаг 2. Найдём высоту $h = CM$ по теореме Пифагора в прямоугольном треугольнике $ACM$: +$$AC^2 = AM^2 + CM^2 \\implies 13^2 = 5^2 + h^2$$ +$$h^2 = 169 - 25 = 144 \\implies h = \\sqrt{144} = 12\\text{ см}$$ +Шаг 3. Считаем площадь: +$$S = \\dfrac{1}{2}\\cdot 10\\cdot 12 = 60\\text{ см}^2$$ +

Альтернативный способ — формула Герона. +
Формула Герона: $S = \\sqrt{p(p-a)(p-b)(p-c)}$, где $p = \\dfrac{a+b+c}{2}$ — полупериметр треугольника со сторонами $a$, $b$, $c$. +
Шаг 1. Полупериметр: $p = \\dfrac{13+13+10}{2} = \\dfrac{36}{2} = 18$ см. +
Шаг 2. Подставляем в формулу: +$$S = \\sqrt{18\\cdot(18-13)\\cdot(18-13)\\cdot(18-10)} = \\sqrt{18\\cdot 5\\cdot 5\\cdot 8} = \\sqrt{3600} = 60\\text{ см}^2$$ +Оба способа дают один и тот же ответ. +
Ответ: $60$ см²
` + }, + { + text: `Упростите выражение $\\sqrt{6}\\cdot(\\sqrt{25} - \\sqrt{96} + 3\\sqrt{6})\\cdot(-\\sqrt{54})$.`, + sol: `Свойство квадратного корня: $\\sqrt{a\\cdot b} = \\sqrt{a}\\cdot\\sqrt{b}$ (для $a,b\\geq 0$), а также $\\sqrt{a}\\cdot\\sqrt{a} = a$ (для $a\\geq 0$). +
Шаг 1. Упрощаем каждый корень в скобках, вынося полные квадраты из-под корня: +$$\\sqrt{25} = 5$$ +$$\\sqrt{96} = \\sqrt{16\\cdot 6} = \\sqrt{16}\\cdot\\sqrt{6} = 4\\sqrt{6}$$ +$$\\sqrt{54} = \\sqrt{9\\cdot 6} = \\sqrt{9}\\cdot\\sqrt{6} = 3\\sqrt{6}$$ +Шаг 2. Подставляем и приводим подобные слагаемые $-4\\sqrt{6}$ и $3\\sqrt{6}$: +$$5 - 4\\sqrt{6} + 3\\sqrt{6} = 5 - \\sqrt{6}$$ +Исходное выражение принимает вид: +$$\\sqrt{6}\\cdot(5 - \\sqrt{6})\\cdot(-3\\sqrt{6})$$ +Шаг 3. Перемножаем крайние множители $\\sqrt{6}$ и $-3\\sqrt{6}$: +$$\\sqrt{6}\\cdot(-3\\sqrt{6}) = -3\\cdot(\\sqrt{6})^2 = -3\\cdot 6 = -18$$ +Шаг 4. Умножаем результат на оставшуюся скобку: +$$-18\\cdot(5 - \\sqrt{6}) = -18\\cdot 5 + 18\\sqrt{6} = -90 + 18\\sqrt{6} = 18(\\sqrt{6} - 5)$$ +
Ответ: $18(\\sqrt{6}-5)$
` + }, + { + text: `Найдите область определения функции + $y = \\dfrac{5}{x^2-9} + \\sqrt{24-8x}$.`, + sol: `Правила нахождения области определения: +
1) Знаменатель дроби не может равняться нулю: $\\dfrac{1}{g(x)}$ определена при $g(x) \\neq 0$. +
2) Подкоренное выражение чётной степени должно быть неотрицательным: $\\sqrt{f(x)}$ определён при $f(x) \\geq 0$. +
В функции есть и дробь, и корень — выписываем оба условия. +
Шаг 1. Знаменатель $x^2 - 9$ должен быть $\\neq 0$: +$$x^2 - 9 \\neq 0 \\implies x^2 \\neq 9 \\implies x \\neq \\pm 3$$ +Шаг 2. Подкоренное выражение $24 - 8x$ должно быть $\\geq 0$: +$$24 - 8x \\geq 0 \\implies 8x \\leq 24 \\implies x \\leq 3$$ +Шаг 3. Объединяем: $x \\leq 3$ и $x \\neq -3$ и $x \\neq 3$. +
Условие $x \\leq 3$ с исключением $x = 3$ даёт $x \\lt 3$. С учётом $x \\neq -3$: +$$x \\in (-\\infty;\\,-3)\\cup(-3;\\,3)$$ +
Ответ: $(-\\infty;\\,-3)\\cup(-3;\\,3)$
` + }, + { + text: `Найдите значение выражения $75x_0$, где $x_0$ — наименьший корень уравнения + $$\\dfrac{x^2}{4x^2-4x+1} - \\dfrac{4x}{2x-1} + 3 = 0.$$`, + sol: `Формула квадрата разности: $(a - b)^2 = a^2 - 2ab + b^2$. +
Шаг 1. Замечаем, что $4x^2 - 4x + 1 = (2x)^2 - 2\\cdot 2x\\cdot 1 + 1^2 = (2x - 1)^2$. +
Уравнение принимает вид: +$$\\dfrac{x^2}{(2x - 1)^2} - \\dfrac{4x}{2x - 1} + 3 = 0$$ +ОДЗ: $2x - 1 \\neq 0$, то есть $x \\neq \\dfrac{1}{2}$. +
Шаг 2. Замена переменной. Заметим, что $\\dfrac{x^2}{(2x - 1)^2} = \\left(\\dfrac{x}{2x - 1}\\right)^2$. +
Пусть $t = \\dfrac{x}{2x - 1}$. Уравнение становится: +$$t^2 - 4t + 3 = 0$$ +Шаг 3. По теореме Виета (сумма $4$, произведение $3$): корни $1$ и $3$. +$$(t - 1)(t - 3) = 0 \\implies t = 1 \\text{ или } t = 3$$ +Шаг 4. Возвращаемся к $x$. +
— При $t = 1$: $\\dfrac{x}{2x - 1} = 1 \\implies x = 2x - 1 \\implies -x = -1 \\implies x = 1$. +
— При $t = 3$: $\\dfrac{x}{2x - 1} = 3 \\implies x = 6x - 3 \\implies -5x = -3 \\implies x = \\dfrac{3}{5}$. +
Шаг 5. Проверка ОДЗ. Оба значения $\\neq \\dfrac{1}{2}$ ✓. +
Шаг 6. Сравним корни: $\\dfrac{3}{5} = 0{,}6 \\lt 1$. Наименьший корень: +$$x_0 = \\dfrac{3}{5}$$ +$$75x_0 = 75\\cdot\\dfrac{3}{5} = \\dfrac{75\\cdot 3}{5} = 15\\cdot 3 = 45$$ +
Ответ: $45$
` + }, + { + text: `Найдите площадь описанной равнобедренной трапеции, если точка касания вписанной + в неё окружности делит боковую сторону на отрезки, равные $2$ см и $8$ см.`, + sol: `Шаг 1. Основания трапеции. +
Боковая сторона $= 2+8 = 10$ см. По свойству касательных от каждой вершины оба касательных отрезка равны. +
От вершин большего основания — по $8$, от вершин меньшего — по $2$. + + + + + + + + + + + + + + + + + + + + + + + A + D + B + C + O + + 8 + 2 + 8 + 2 + + AD = 16 + BC = 4 + + r = 4 + +$$AD = 8+8 = 16\\text{ см}, \\quad BC = 2+2 = 4\\text{ см}$$ +Шаг 2. Высота трапеции. +
Горизонтальный выступ ноги: $\\dfrac{16-4}{2} = 6$ см. +$$h = \\sqrt{10^2-6^2} = \\sqrt{100-36} = \\sqrt{64} = 8\\text{ см}$$ +Шаг 3. Площадь. +$$S = \\dfrac{AD+BC}{2}\\cdot h = \\dfrac{16+4}{2}\\cdot8 = 10\\cdot8 = 80\\text{ см}^2$$ +
Ответ: $80$ см²
` + }, + { + text: `В компанию поступил заказ на укладку $175$ м² напольной плитки. + Плиточник принял решение укладывать на $10$ м² в день больше, чем запланировал ранее. + В результате работа была закончена на $2$ дня раньше установленного срока. + Успеет ли плиточник выполнить заказ за $7$ рабочих дней, если будет работать + по первоначальному плану? Ответ обоснуйте.`, + sol: `Пусть плановая выработка $= x$ м²/день. +
По плану: $\\dfrac{175}{x}$ дней. С ускорением: $\\dfrac{175}{x+10}$ дней, на $2$ меньше. +$$\\dfrac{175}{x} - \\dfrac{175}{x+10} = 2$$ +Умножаем на $x(x+10)$: +$$175(x+10) - 175x = 2x(x+10)$$ +$$1750 = 2x^2+20x \\implies x^2+10x-875=0$$ +$$D = 100+3500 = 3600 = 60^2 \\implies x = \\dfrac{-10+60}{2} = 25\\text{ м}^2/\\text{день}$$ +Плановый срок: $\\dfrac{175}{25} = 7$ дней. +
Проверка: при $35$ м²/день: $175\\div35=5$ дней, $7-5=2$ ✓ +
Ответ на вопрос: за $7$ дней при выработке $25$ м²/день плиточник уложит $7\\times25=175$ м². +
Ответ: да, успеет — уложит ровно $175$ м² за $7$ дней
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v45.js b/frontend/js/exam9/variants/v45.js new file mode 100644 index 0000000..b016fd4 --- /dev/null +++ b/frontend/js/exam9/variants/v45.js @@ -0,0 +1,187 @@ +VARIANTS[45] = { + label: "Вариант 45", + tasks: [ + { + text: `Определите наименьшее натуральное число, принадлежащее промежутку + $\\left(-\\dfrac{3}{4};\\; 5{,}6\\right)$:`, + opts: [ + ["а", "$-1$"], ["б", "$0$"], ["в", "$1$"], ["г", "$2$"], ["д", "$5$"], + ], + sol: `Натуральные числа: $1, 2, 3, \\ldots$ Наименьшее из них, попадающее в промежуток $(-0{,}75;\\;5{,}6)$, — это $1$. +
Ответ: в) $1$
` + }, + { + text: `График обратной пропорциональности $y = \\dfrac{k}{x}$ проходит через точку + $(-\\sqrt{3};\\; 4\\sqrt{3})$. Коэффициент $k$ равен:`, + opts: [ + ["а", "$k = -\\sqrt{3}$"], ["б", "$k = 4\\sqrt{3}$"], ["в", "$k = -12$"], + ["г", "$k = 4$"], ["д", "$k = 3$"], + ], + sol: `$k = x\\cdot y = (-\\sqrt{3})\\cdot4\\sqrt{3} = -4\\cdot3 = -12$ +
Ответ: в) $k=-12$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "диагонали ромба взаимно перпендикулярны;"], + ["б", "если две стороны одного треугольника соответственно равны двум сторонам другого, то такие треугольники равны;"], + ["в", "хорда окружности, проходящая через её центр, является диаметром;"], + ["г", "боковые стороны равнобедренной трапеции равны между собой?"], + ], + sol: `
    +
  • а) Диагонали ромба перпендикулярны — верно
    • +
  • б) «Две стороны равны ⟹ треугольники равны» — НЕВЕРНО. Для равенства треугольников нужно ССС, СУС, УСУ или УУС. Два равных основания — недостаточно.
    • +
  • в) Хорда через центр = диаметр — верно
    • +
  • г) Боковые стороны равнобедренной трапеции равны — верно
    • +
+
Ответ: б)
` + }, + { + text: `Решите уравнение $3x^2 + x = 0$. + В ответ запишите среднее арифметическое корней уравнения.`, + sol: `$$x(3x+1)=0 \\implies x_1=0,\\quad x_2=-\\dfrac{1}{3}$$ +Среднее арифметическое: +$$\\dfrac{x_1+x_2}{2}=\\dfrac{0+\\left(-\\dfrac{1}{3}\\right)}{2}=-\\dfrac{1}{6}$$ +
Ответ: $-\\dfrac{1}{6}$
` + }, + { + text: `$ABCD$ — прямоугольник, $O$ — точка пересечения его диагоналей. + Угол $AOB$ равен $46^{\\circ}$. Найдите угол $ADB$.`, + sol: ` + + + + + + + + A + B + C + D + O + 46° + 23° + +В прямоугольнике диагонали равны и точкой пересечения делятся пополам: $OA = OB = OC = OD$. +
Треугольник $AOB$ — равнобедренный ($OA=OB$): +$$\\angle OAB = \\angle OBA = \\dfrac{180°-46°}{2} = 67°$$ +Углы при $O$: $A,O,C$ лежат на диагонали $AC$, поэтому $\\angle DOA = 180°-46° = 134°$. +
Треугольник $DOA$ — равнобедренный ($OD=OA$): +$$\\angle ODA = \\dfrac{180°-134°}{2} = 23°$$ +Так как $O$ лежит на отрезке $BD$, то $\\angle ADB = \\angle ODA = 23°$. +
Ответ: $\\angle ADB = 23°$
` + }, + { + text: `Найдите значение выражения $14A$, если + $A = (2\\sqrt{2} - 1)(\\sqrt{8} + 1) - 8 \\cdot \\dfrac{1}{7}$.`, + sol: `Формула разности квадратов: $(a-b)(a+b) = a^2 - b^2$. +
Шаг 1. Упрощаем $\\sqrt{8}$, вынося полный квадрат из-под корня: +$$\\sqrt{8} = \\sqrt{4\\cdot 2} = \\sqrt{4}\\cdot\\sqrt{2} = 2\\sqrt{2}$$ +Значит, первое произведение принимает вид $(2\\sqrt{2} - 1)(2\\sqrt{2} + 1)$. +
Шаг 2. Применяем формулу разности квадратов (здесь $a = 2\\sqrt{2}$, $b = 1$): +$$(2\\sqrt{2} - 1)(2\\sqrt{2} + 1) = (2\\sqrt{2})^2 - 1^2 = 4\\cdot 2 - 1 = 8 - 1 = 7$$ +Шаг 3. Подставляем в выражение для $A$: +$$A = 7 - 8\\cdot\\dfrac{1}{7} = 7 - \\dfrac{8}{7} = \\dfrac{7\\cdot 7 - 8}{7} = \\dfrac{49 - 8}{7} = \\dfrac{41}{7}$$ +Шаг 4. Находим $14A$: +$$14A = 14\\cdot\\dfrac{41}{7} = \\dfrac{14}{7}\\cdot 41 = 2\\cdot 41 = 82$$ +
Ответ: $82$
` + }, + { + text: `При каких целых отрицательных значениях $n$ верно неравенство + $\\dfrac{n+1}{3} - \\dfrac{n+2}{6} < \\dfrac{n+3}{2}$?`, + sol: `Свойство неравенства: при умножении обеих частей неравенства на одно и то же положительное число знак неравенства сохраняется. +
Шаг 1. Наименьший общий знаменатель дробей $3,\\,6,\\,2$ равен $6$. Умножаем обе части на $6$: +$$6\\cdot\\dfrac{n+1}{3} - 6\\cdot\\dfrac{n+2}{6} \\lt 6\\cdot\\dfrac{n+3}{2}$$ +$$2(n + 1) - (n + 2) \\lt 3(n + 3)$$ +Шаг 2. Раскрываем скобки: +$$2n + 2 - n - 2 \\lt 3n + 9$$ +$$n \\lt 3n + 9$$ +Шаг 3. Переносим $3n$ влево, а $n$ — вправо: +$$n - 3n \\lt 9 \\implies -2n \\lt 9$$ +Шаг 4. Делим обе части на $-2$. Важно: при делении на отрицательное число знак неравенства меняется на противоположный: +$$n \\gt -\\dfrac{9}{2} = -4{,}5$$ +Шаг 5. Выбираем целые отрицательные числа, большие $-4{,}5$: +$$n \\in \\{-4,\\,-3,\\,-2,\\,-1\\}$$ +
Ответ: $n\\in\\{-4,\\;-3,\\;-2,\\;-1\\}$
` + }, + { + text: `Найдите значение выражения $6(x - y)$, где $(x;\\; y)$ — решение системы уравнений + $$\\begin{cases} 3x + y = 2, \\\\[4pt] y^2 + 6xy + 9x^2 = -y. \\end{cases}$$`, + sol: `Формула квадрата суммы: $(a + b)^2 = a^2 + 2ab + b^2$. +
Шаг 1. Замечаем структуру левой части второго уравнения: +$$y^2 + 6xy + 9x^2 = y^2 + 2\\cdot y\\cdot 3x + (3x)^2 = (y + 3x)^2 = (3x + y)^2$$ +Шаг 2. Из первого уравнения системы: $3x + y = 2$. Подставляем во второе: +$$(3x + y)^2 = -y \\implies 2^2 = -y \\implies 4 = -y$$ +$$y = -4$$ +Шаг 3. Подставляем $y = -4$ в $3x + y = 2$: +$$3x + (-4) = 2 \\implies 3x = 6 \\implies x = 2$$ +Проверка: подставим в исходное второе уравнение: +$$(-4)^2 + 6\\cdot 2\\cdot(-4) + 9\\cdot 2^2 = 16 - 48 + 36 = 4$$ +$$-y = -(-4) = 4 \\checkmark$$ +Шаг 4. Считаем искомое выражение: +$$6(x - y) = 6\\cdot(2 - (-4)) = 6\\cdot 6 = 36$$ +
Ответ: $36$
` + }, + { + text: `При открытии торгов в среду акции компании подорожали на некоторое количество процентов, + а в четверг — подешевели на то же количество процентов. + В результате они стали стоить на $4\\%$ дешевле, чем при открытии торгов в среду. + На сколько процентов подорожали акции в среду?`, + sol: `Метод процентных коэффициентов: увеличение величины на $p\\%$ соответствует умножению на $\\left(1 + \\dfrac{p}{100}\\right)$, уменьшение на $p\\%$ — умножению на $\\left(1 - \\dfrac{p}{100}\\right)$. Здесь же применяется формула разности квадратов: $(1+a)(1-a) = 1 - a^2$. +
Шаг 1. Пусть $P$ — начальная цена акции, а $p$ — искомый процент изменения. +
Шаг 2. В среду цена выросла на $p\\%$, значит к концу среды стала равна: +$$P_1 = P\\cdot\\left(1 + \\dfrac{p}{100}\\right)$$ +Шаг 3. В четверг цена снизилась на $p\\%$ от новой цены $P_1$, значит к концу четверга: +$$P_2 = P_1\\cdot\\left(1 - \\dfrac{p}{100}\\right) = P\\cdot\\left(1 + \\dfrac{p}{100}\\right)\\left(1 - \\dfrac{p}{100}\\right)$$ +По формуле разности квадратов: +$$P_2 = P\\cdot\\left(1 - \\dfrac{p^2}{10000}\\right)$$ +Шаг 4. По условию итоговая цена на $4\\%$ ниже начальной, то есть $P_2 = 0{,}96\\cdot P$. Получаем: +$$1 - \\dfrac{p^2}{10000} = 0{,}96$$ +Шаг 5. Решаем уравнение: +$$\\dfrac{p^2}{10000} = 0{,}04 \\implies p^2 = 400 \\implies p = 20$$ +(берём положительный корень, так как $p$ — процент роста). +
Ответ: подорожали на $20\\%$
` + }, + { + text: `$ABCD$ — вписанная трапеция. Центр $O$ описанной окружности лежит на большем основании $AD$, + $CH$ — высота трапеции. Найдите площадь трапеции, если $AC = 10$ см, $HD = 4{,}5$ см.`, + sol: `Шаг 1. AD — диаметр. +
Раз центр $O$ лежит на хорде $AD$, значит $AD$ проходит через центр — $AD$ является диаметром. +
По теореме Фалеса: $\\angle ACD = 90°$ (вписанный угол, опирающийся на диаметр). + + + + + + + + + A + D + B + C + H + O + 10 + h + 4,5 + BC + +Шаг 2. Находим $AD$. +
Из прямоугольного $\\triangle ACD$ (прямой угол при $C$): $AC^2 = AH\\cdot AD$ (свойство высоты прямоугольного треугольника). +$$AH = AD - HD = AD - 4{,}5$$ +$$10^2 = (AD-4{,}5)\\cdot AD \\implies AD^2 - 4{,}5\\,AD - 100 = 0$$ +$$\\times2:\\quad 2AD^2-9\\,AD-200=0, \\quad D=81+1600=1681=41^2$$ +$$AD=\\dfrac{9+41}{4}=\\dfrac{50}{4}=12{,}5\\text{ см}$$ +Шаг 3. Высота $CH$. +$$AH = 12{,}5-4{,}5=8, \\quad CH^2=AH\\cdot HD = 8\\cdot4{,}5=36 \\implies CH=6$$ +Шаг 4. Основание $BC$. +
Трапеция равнобедренная (вписанная). По симметрии: расстояние от $B$ до $AD$ = $AH'=4{,}5$ (зеркально). +$$BC = AD - 2\\cdot HD = 12{,}5 - 2\\cdot4{,}5 = 3{,}5\\text{ см}$$ +Шаг 5. Площадь. +$$S = \\dfrac{AD+BC}{2}\\cdot CH = \\dfrac{12{,}5+3{,}5}{2}\\cdot6 = 8\\cdot6 = 48\\text{ см}^2$$ +
Ответ: $48$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v46.js b/frontend/js/exam9/variants/v46.js new file mode 100644 index 0000000..01a8374 --- /dev/null +++ b/frontend/js/exam9/variants/v46.js @@ -0,0 +1,199 @@ +VARIANTS[46] = { + label: "Вариант 46", + tasks: [ + { + text: `Определите наибольшее натуральное число, принадлежащее промежутку + $\\left(-\\dfrac{2}{3};\\; 7{,}1\\right)$:`, + opts: [ + ["а", "$2$"], ["б", "$1$"], ["в", "$0$"], ["г", "$6$"], ["д", "$7$"], + ], + sol: `Натуральные числа: $1, 2, 3, \\ldots$ Все они принадлежат промежутку $(-0{,}67;\\;7{,}1)$, если не превышают $7$. Наибольшее такое число — $7$, так как $7 \\lt 7{,}1$, а $8 \\gt 7{,}1$. +
Ответ: д) $7$
` + }, + { + text: `График обратной пропорциональности $y = \\dfrac{k}{x}$ проходит через точку + $(\\sqrt{5};\\; -2\\sqrt{5})$. Коэффициент $k$ равен:`, + opts: [ + ["а", "$k = \\sqrt{5}$"], ["б", "$k = -2\\sqrt{5}$"], ["в", "$k = -10$"], + ["г", "$k = 2$"], ["д", "$k = -5$"], + ], + sol: `$k = x\\cdot y = \\sqrt{5}\\cdot(-2\\sqrt{5}) = -2\\cdot5 = -10$ +
Ответ: в) $k=-10$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "диагонали ромба лежат на биссектрисах его углов;"], + ["б", "диаметр окружности равен двум радиусам;"], + ["в", "если два угла одного треугольника соответственно равны двум углам другого, то такие треугольники равны;"], + ["г", "прямоугольная трапеция имеет два прямых угла?"], + ], + sol: `
    +
  • а) Диагонали ромба лежат на биссектрисах его углов — верно
    • +
  • б) Диаметр равен двум радиусам — верно
    • +
  • в) «Два равных угла ⟹ треугольники равны» — НЕВЕРНО. Равенство двух углов означает лишь подобие (ААА), но не равенство: стороны могут различаться.
    • +
  • г) Прямоугольная трапеция имеет два прямых угла — верно
    • +
+
Ответ: в)
` + }, + { + text: `Решите уравнение $2x^2 + x = 0$. + В ответ запишите среднее арифметическое корней уравнения.`, + sol: `$$x(2x+1)=0 \\implies x_1=0,\\quad x_2=-\\dfrac{1}{2}$$ +Среднее арифметическое: +$$\\dfrac{x_1+x_2}{2}=\\dfrac{0+\\left(-\\dfrac{1}{2}\\right)}{2}=-\\dfrac{1}{4}$$ +
Ответ: $-\\dfrac{1}{4}$
` + }, + { + text: `$ABCD$ — прямоугольник, $O$ — точка пересечения его диагоналей. + Угол $DBC$ равен $32^{\\circ}$. Найдите угол $AOD$.`, + sol: ` + + + + + + + + A + B + C + D + O + 32° + 116° + +Свойства прямоугольника: +
1) Все углы прямоугольника прямые ($90°$). +
2) Диагонали прямоугольника равны и точкой пересечения делятся пополам. +
Шаг 1. В прямоугольнике $\\angle ABC = 90°$. Точка $O$ лежит внутри угла $ABC$, поэтому: +$$\\angle ABD = \\angle ABC - \\angle DBC = 90° - 32° = 58°$$ +Шаг 2. По свойству диагоналей: $OA = OB$ (половинки равных диагоналей). +
Значит, треугольник $AOB$ — равнобедренный с основанием $AB$. По свойству равнобедренного треугольника углы при основании равны: +$$\\angle OAB = \\angle OBA = \\angle ABD = 58°$$ +Шаг 3. По теореме о сумме углов треугольника (сумма $= 180°$): +$$\\angle AOB = 180° - 58° - 58° = 64°$$ +Шаг 4. Точки $A$, $O$, $C$ лежат на одной прямой (диагональ $AC$), поэтому углы $\\angle AOB$ и $\\angle BOC$ — смежные (как и $\\angle AOD$ и $\\angle DOC$). Углы $\\angle AOD$ и $\\angle AOB$ смежные: +$$\\angle AOD = 180° - \\angle AOB = 180° - 64° = 116°$$ +
Ответ: $\\angle AOD = 116°$
` + }, + { + text: `Найдите значение выражения $56A$, если + $A = (3\\sqrt{2} - 2)(\\sqrt{18} + 2) - 14 \\cdot \\dfrac{1}{8}$.`, + sol: `Формула разности квадратов: $(a-b)(a+b) = a^2 - b^2$. +
Шаг 1. Упрощаем $\\sqrt{18}$, вынося полный квадрат: +$$\\sqrt{18} = \\sqrt{9\\cdot 2} = \\sqrt{9}\\cdot\\sqrt{2} = 3\\sqrt{2}$$ +Значит, первое произведение принимает вид $(3\\sqrt{2} - 2)(3\\sqrt{2} + 2)$. +
Шаг 2. Применяем формулу разности квадратов (здесь $a = 3\\sqrt{2}$, $b = 2$): +$$(3\\sqrt{2} - 2)(3\\sqrt{2} + 2) = (3\\sqrt{2})^2 - 2^2 = 9\\cdot 2 - 4 = 18 - 4 = 14$$ +Шаг 3. Подставляем в выражение для $A$: +$$A = 14 - 14\\cdot\\dfrac{1}{8} = 14 - \\dfrac{14}{8} = 14 - \\dfrac{7}{4} = \\dfrac{56 - 7}{4} = \\dfrac{49}{4}$$ +Шаг 4. Находим $56A$: +$$56A = 56\\cdot\\dfrac{49}{4} = \\dfrac{56}{4}\\cdot 49 = 14\\cdot 49 = 686$$ +
Ответ: $686$
` + }, + { + text: `При каких натуральных значениях $m$ верно неравенство + $\\dfrac{m+1}{2} - \\dfrac{m-2}{3} > \\dfrac{m+3}{4}$?`, + sol: `Свойство неравенства: при умножении обеих частей на положительное число знак неравенства сохраняется. +
Шаг 1. Наименьший общий знаменатель дробей $2,\\,3,\\,4$ равен $12$. Умножаем обе части на $12$: +$$12\\cdot\\dfrac{m+1}{2} - 12\\cdot\\dfrac{m-2}{3} \\gt 12\\cdot\\dfrac{m+3}{4}$$ +$$6(m + 1) - 4(m - 2) \\gt 3(m + 3)$$ +Шаг 2. Раскрываем скобки: +$$6m + 6 - 4m + 8 \\gt 3m + 9$$ +$$2m + 14 \\gt 3m + 9$$ +Шаг 3. Переносим $3m$ влево, числа — вправо: +$$2m - 3m \\gt 9 - 14 \\implies -m \\gt -5$$ +Шаг 4. Умножаем на $-1$. Важно: при умножении на отрицательное число знак неравенства меняется на противоположный: +$$m \\lt 5$$ +Шаг 5. Натуральные числа, меньшие $5$: +$$m \\in \\{1,\\,2,\\,3,\\,4\\}$$ +
Ответ: $m\\in\\{1,\\;2,\\;3,\\;4\\}$
` + }, + { + text: `Найдите значение выражения $10(x - y)$, где $(x;\\; y)$ — решение системы уравнений + $$\\begin{cases} x^2 + 4xy + 4y^2 = -x - 6y, \\\\[4pt] x + 2y = 1. \\end{cases}$$`, + sol: `Формула квадрата суммы: $(a + b)^2 = a^2 + 2ab + b^2$. +
Шаг 1. Замечаем структуру левой части первого уравнения: +$$x^2 + 4xy + 4y^2 = x^2 + 2\\cdot x\\cdot 2y + (2y)^2 = (x + 2y)^2$$ +Шаг 2. Из второго уравнения системы: $x + 2y = 1$. Подставляем в первое: +$$(x + 2y)^2 = -x - 6y$$ +$$1^2 = -(x + 6y) \\implies x + 6y = -1$$ +Шаг 3. Получили новую систему: $\\{x + 2y = 1;\\; x + 6y = -1\\}$. +
Вычтем первое уравнение из второго (метод вычитания исключает $x$): +$$(x + 6y) - (x + 2y) = -1 - 1 \\implies 4y = -2 \\implies y = -\\dfrac{1}{2}$$ +Шаг 4. Подставляем $y = -\\dfrac{1}{2}$ в $x + 2y = 1$: +$$x + 2\\cdot\\left(-\\dfrac{1}{2}\\right) = 1 \\implies x - 1 = 1 \\implies x = 2$$ +Шаг 5. Вычисляем искомое выражение: +$$10(x - y) = 10\\cdot\\left(2 - \\left(-\\dfrac{1}{2}\\right)\\right) = 10\\cdot\\dfrac{5}{2} = 25$$ +
Ответ: $25$
` + }, + { + text: `При открытии торгов в среду акции компании подешевели на некоторое количество процентов, + а в четверг — подорожали на то же количество процентов. + В результате они стали стоить на $9\\%$ дешевле, чем при открытии торгов в среду. + На сколько процентов подорожали акции в четверг?`, + sol: `Метод процентных коэффициентов: уменьшение на $p\\%$ соответствует умножению на $\\left(1 - \\dfrac{p}{100}\\right)$, увеличение на $p\\%$ — на $\\left(1 + \\dfrac{p}{100}\\right)$. Также используется формула разности квадратов: $(1-a)(1+a) = 1 - a^2$. +
Шаг 1. Пусть $P$ — цена при открытии торгов в среду, а $p$ — искомый процент. +
Шаг 2. В среду цена снизилась на $p\\%$, значит к концу среды: +$$P_1 = P\\cdot\\left(1 - \\dfrac{p}{100}\\right)$$ +Шаг 3. В четверг цена выросла на $p\\%$ от $P_1$, значит к концу четверга: +$$P_2 = P_1\\cdot\\left(1 + \\dfrac{p}{100}\\right) = P\\cdot\\left(1 - \\dfrac{p}{100}\\right)\\left(1 + \\dfrac{p}{100}\\right)$$ +По формуле разности квадратов: +$$P_2 = P\\cdot\\left(1 - \\dfrac{p^2}{10000}\\right)$$ +Шаг 4. По условию итоговая цена на $9\\%$ ниже начальной, то есть $P_2 = 0{,}91\\cdot P$: +$$1 - \\dfrac{p^2}{10000} = 0{,}91$$ +Шаг 5. Решаем: +$$\\dfrac{p^2}{10000} = 0{,}09 \\implies p^2 = 900 \\implies p = 30$$ +(берём положительный корень). +
Ответ: подорожали на $30\\%$
` + }, + { + text: `$ABCD$ — вписанная трапеция. Центр $O$ описанной окружности лежит на большем основании $AD$, + $BH$ — высота трапеции. Найдите площадь трапеции, если $BD = 20$ см, $AH = 9$ см.`, + sol: `Теорема Фалеса (о вписанном угле, опирающемся на диаметр): вписанный угол, опирающийся на диаметр, — прямой ($90°$). +
Свойство высоты прямоугольного треугольника: высота, проведённая из вершины прямого угла к гипотенузе, удовлетворяет соотношениям: +
$h^2 = m\\cdot n$ (где $m$, $n$ — проекции катетов на гипотенузу), а также $a^2 = m\\cdot c$, $b^2 = n\\cdot c$, где $a$, $b$ — катеты, $c$ — гипотенуза. +
Шаг 1. Так как центр $O$ описанной окружности лежит на хорде $AD$, то $AD$ проходит через центр, то есть $AD$ — диаметр. +
По теореме Фалеса вписанный угол $\\angle ABD = 90°$ (опирается на диаметр $AD$). + + + + + + + + + + + + A + D + B + C + H + O + BD=20 + h + AH=9 + BC + +Шаг 2. Находим $AD$. В прямоугольном $\\triangle ABD$ (прямой угол при $B$) $BH$ — высота, опущенная на гипотенузу $AD$. +
По свойству высоты: $BD^2 = HD\\cdot AD$, где $HD$ — проекция катета $BD$ на гипотенузу. +$$20^2 = HD\\cdot AD \\implies 400 = HD\\cdot AD$$ +Также $HD = AD - AH = AD - 9$. Подставляем: +$$400 = (AD - 9)\\cdot AD \\implies AD^2 - 9AD - 400 = 0$$ +Шаг 3. Решаем по формуле дискриминанта: +$$D = 81 + 1600 = 1681 = 41^2 \\implies AD = \\dfrac{9 + 41}{2} = 25\\text{ см}$$ +(второй корень отрицательный, не подходит). +
Шаг 4. Находим $HD$, $BH$ и второе основание трапеции. +$$HD = 25 - 9 = 16\\text{ см}$$ +По свойству высоты $BH^2 = AH\\cdot HD = 9\\cdot 16 = 144$, значит $BH = 12$ см. +
Трапеция $ABCD$ равнобедренная (как вписанная). По симметрии расстояние от $C$ до $AD$ тоже даёт «выступ» $9$ см справа. Тогда: +$$BC = AD - 2\\cdot AH = 25 - 2\\cdot 9 = 7\\text{ см}$$ +Шаг 5. По формуле площади трапеции: +$$S = \\dfrac{AD + BC}{2}\\cdot BH = \\dfrac{25 + 7}{2}\\cdot 12 = 16\\cdot 12 = 192\\text{ см}^2$$ +
Ответ: $192$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v47.js b/frontend/js/exam9/variants/v47.js new file mode 100644 index 0000000..475b753 --- /dev/null +++ b/frontend/js/exam9/variants/v47.js @@ -0,0 +1,204 @@ +VARIANTS[47] = { + label: "Вариант 47", + tasks: [ + { + text: `Определите, какое из данных равенств является верным:`, + opts: [ + ["а", "$\\sqrt{72} = 36\\sqrt{2}$"], ["б", "$\\sqrt{72} = 2\\sqrt{6}$"], ["в", "$\\sqrt{72} = 6\\sqrt{2}$"], + ["г", "$\\sqrt{72} = 12\\sqrt{2}$"], ["д", "$\\sqrt{72} = 24\\sqrt{2}$"], + ], + sol: `Разложим подкоренное число так, чтобы выделить полный квадрат: + $$\\sqrt{72} = \\sqrt{36 \\cdot 2} = \\sqrt{36}\\cdot\\sqrt{2} = 6\\sqrt{2}.$$ + Проверим остальные варианты:
+ $36\\sqrt{2}\\approx 50{,}9$, $2\\sqrt{6}\\approx 4{,}9$, $12\\sqrt{2}\\approx 17$, $24\\sqrt{2}\\approx 33{,}9$, + а $\\sqrt{72}\\approx 8{,}49$. Совпадает только $6\\sqrt{2}$. +
Ответ: в) $\\sqrt{72} = 6\\sqrt{2}$.
` + }, + { + text: `Значение выражения $\\dfrac{6^4}{6^2} + 6^1$ равно:`, + opts: [ + ["а", "$36$"], ["б", "$37$"], ["в", "$18$"], ["г", "$42$"], ["д", "$48$"], + ], + sol: `Используем свойство степеней $\\dfrac{a^m}{a^n}=a^{m-n}$: + $$\\dfrac{6^4}{6^2} + 6^1 = 6^{4-2} + 6 = 6^2 + 6 = 36 + 6 = 42.$$ +
Ответ: г) $42$.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "диагонали любого прямоугольника взаимно перпендикулярны;"], + ["б", "высота ромба равна диаметру вписанной в него окружности;"], + ["в", "центр окружности, описанной около треугольника, лежит на пересечении серединных перпендикуляров к сторонам треугольника;"], + ["г", "угол, равный $89^{\\circ}$, — острый?"], + ], + sol: `Проанализируем каждое утверждение: +
    +
  • а) Диагонали прямоугольника равны и точкой пересечения делятся пополам, + но взаимно перпендикулярны они только в частном случае — в квадрате. + В произвольном прямоугольнике это неверно.
    • +
  • б) Верно: высота ромба равна диаметру вписанной окружности (стандартное свойство).
    • +
  • в) Верно: центр описанной окружности — точка пересечения серединных перпендикуляров.
    • +
  • г) Верно: угол меньше $90^{\\circ}$ — острый, $89^{\\circ}<90^{\\circ}$.
    • +
+
Ответ: а) — утверждение неверно.
` + }, + { + text: `При каких значениях переменной $x$ равны значения трёхчленов + $5x^2 - 3x + 4$ и $3x + 3 - 4x^2$?`, + sol: `Приравняем трёхчлены: + $$5x^2 - 3x + 4 = 3x + 3 - 4x^2.$$ + Перенесём всё в левую часть: + $$5x^2 - 3x + 4 - 3x - 3 + 4x^2 = 0,$$ + $$9x^2 - 6x + 1 = 0.$$ + Замечаем полный квадрат: $9x^2 - 6x + 1 = (3x-1)^2$. Значит, + $$(3x-1)^2 = 0 \\;\\Longrightarrow\\; 3x - 1 = 0 \\;\\Longrightarrow\\; x = \\dfrac{1}{3}.$$ +
Ответ: $x = \\dfrac{1}{3}$.
` + }, + { + text: `В прямоугольном треугольнике $ABC$ $\\angle B = 90^{\\circ}$, $BC = 20$ см, высота $BH = 12$ см. + Найдите синус угла $A$.`, + sol: ` + + + + + + + + + A + C + B + H + 15 + 20 + 25 + 12 + + В прямоугольном треугольнике $ABC$ ($\\angle B=90^{\\circ}$) $BH$ — высота, проведённая к гипотенузе $AC$. + Запишем площадь двумя способами: + $$S = \\tfrac{1}{2}\\cdot AB\\cdot BC = \\tfrac{1}{2}\\cdot AC\\cdot BH.$$ + Отсюда $AB\\cdot 20 = AC\\cdot 12$, то есть $AC = \\dfrac{5\\,AB}{3}$.
+ По теореме Пифагора $AB^2 + BC^2 = AC^2$: + $$AB^2 + 400 = \\dfrac{25\\,AB^2}{9} \\;\\Longrightarrow\\; \\dfrac{16\\,AB^2}{9} = 400 + \\;\\Longrightarrow\\; AB^2 = 225 \\;\\Longrightarrow\\; AB = 15\\text{ см}.$$ + Тогда $AC = \\dfrac{5\\cdot 15}{3} = 25$ см.
+ Синус угла $A$ — отношение противолежащего катета $BC$ к гипотенузе $AC$: + $$\\sin A = \\dfrac{BC}{AC} = \\dfrac{20}{25} = \\dfrac{4}{5} = 0{,}8.$$ +
Ответ: $\\sin A = \\dfrac{4}{5} = 0{,}8$.
` + }, + { + text: `Упростите выражение + $\\dfrac{5x+6}{x^2-4} - \\dfrac{x}{x^2-4} : \\dfrac{x}{x-2} - \\dfrac{x+2}{x-2}$.`, + sol: `Порядок действий: в выражении без скобок сначала выполняется деление и умножение, а затем сложение и вычитание (слева направо). Также применяется формула разности квадратов: $a^2 - b^2 = (a-b)(a+b)$. +
Шаг 1. Раскладываем $x^2 - 4$ по формуле разности квадратов: +$$x^2 - 4 = (x - 2)(x + 2)$$ +ОДЗ: $x \\neq 2$, $x \\neq -2$, $x \\neq 0$ (так как в выражении есть деление на $\\dfrac{x}{x-2}$). +
Шаг 2. По порядку действий сначала выполняем деление $\\dfrac{x}{x^2-4} : \\dfrac{x}{x-2}$. Делим на дробь — умножаем на обратную: +$$\\dfrac{x}{(x-2)(x+2)} \\cdot \\dfrac{x-2}{x} = \\dfrac{x\\cdot(x-2)}{(x-2)(x+2)\\cdot x} = \\dfrac{1}{x+2}$$ +Шаг 3. Исходное выражение принимает вид: +$$\\dfrac{5x+6}{(x-2)(x+2)} - \\dfrac{1}{x+2} - \\dfrac{x+2}{x-2}$$ +Шаг 4. Приводим к общему знаменателю $(x-2)(x+2)$. Домножаем числители: первой дроби — на $1$, второй — на $(x-2)$, третьей — на $(x+2)$: +$$\\dfrac{(5x+6) - (x-2) - (x+2)^2}{(x-2)(x+2)}$$ +Шаг 5. Раскрываем скобки в числителе, применяя формулу квадрата суммы $(x+2)^2 = x^2 + 4x + 4$: +$$5x + 6 - x + 2 - (x^2 + 4x + 4) = (5x - x - 4x) + (6 + 2 - 4) - x^2 = 0 + 4 - x^2 = 4 - x^2$$ +Шаг 6. Получаем: +$$\\dfrac{4 - x^2}{x^2 - 4} = \\dfrac{-(x^2 - 4)}{x^2 - 4} = -1$$ +
Ответ: $-1$ (при $x \\neq \\pm 2$ и $x \\neq 0$).
` + }, + { + text: `График функции $f(x) = a(x-m)^2 + n$ изображён на рисунке. + Используя график функции, найдите $a$, $m$ и $n$. + Запишите формулу функции $y = f(x)$ в виде многочлена.`, + sol: `Функция $f(x)=a(x-m)^2+n$ — парабола с вершиной в точке $(m;\\,n)$; + знак $a$ определяет направление ветвей ($a>0$ — вверх, $a<0$ — вниз), + а $|a|$ — «крутизну».

+ Алгоритм по графику: +
    +
  • находим координаты вершины параболы — это $m$ (абсцисса) и $n$ (ордината);
    • +
  • берём любую другую точку $(x_0;\\,y_0)$ графика и подставляем в формулу: + $a = \\dfrac{y_0 - n}{(x_0 - m)^2}$;
    • +
  • раскрываем скобки: $f(x)=a(x-m)^2+n = ax^2 - 2am\\,x + (am^2+n)$.
    • +
+ Типичный вариант (вершина $(1;\\,-4)$, ветви вниз, через точку $(0;-5)$): + $m=1$, $n=-4$, $a=\\dfrac{-5-(-4)}{(0-1)^2} = -1$. + $$f(x) = -(x-1)^2 - 4 = -x^2 + 2x - 1 - 4 = -x^2 + 2x - 5.$$ +
Ответ: $a$, $m$, $n$ снимаются с графика по вершине $(m;n)$ и контрольной точке; + $f(x)=ax^2-2am\\,x+(am^2+n)$. Например, при вершине $(1;-4)$ и $a=-1$: $f(x)=-x^2+2x-5$.
` + }, + { + text: `Найдите сумму целых решений системы неравенств + $$\\begin{cases} 9 - 4x < 0, \\\\[4pt] x^2 - 5x \\leq -4. \\end{cases}$$`, + sol: `1) Первое неравенство: + $$9 - 4x < 0 \\;\\Longrightarrow\\; 4x > 9 \\;\\Longrightarrow\\; x > 2{,}25.$$ + 2) Второе неравенство: + $$x^2 - 5x + 4 \\leq 0.$$ + Корни квадратного трёхчлена: $x_{1,2} = \\dfrac{5\\pm\\sqrt{25-16}}{2} = \\dfrac{5\\pm 3}{2}$, + то есть $x_1=1$, $x_2=4$. Так как ветви параболы $y=x^2-5x+4$ направлены вверх, + неравенство $\\leq 0$ выполняется между корнями: $1 \\leq x \\leq 4$.
+ 3) Пересечение: $x>2{,}25$ и $1\\leq x\\leq 4$ дают $2{,}25 < x \\leq 4$.
+ 4) Целые решения: $x=3$ и $x=4$.
+ Сумма: $3+4=7$. +
Ответ: $7$.
` + }, + { + text: `Для перевозки партии щебня массой $1008$ т фирма использует самосвал МАЗ-5551. + По плану норма перевозки ежедневно должна увеличиваться на одно и то же число тонн. + Известно, что за первый день было перевезено $40$ т щебня. + Определите, сколько тонн щебня было перевезено за девятый день, + если вся работа была выполнена за $12$ дней.`, + sol: `Метод арифметической прогрессии. По условию норма ежедневно увеличивается на одно и то же число тонн, значит дневные объёмы образуют арифметическую прогрессию. +
Формулы арифметической прогрессии: +
— $n$-й член: $a_n = a_1 + (n-1)d$, где $d$ — разность прогрессии. +
— Сумма первых $n$ членов: $S_n = \\dfrac{2a_1 + (n-1)d}{2}\\cdot n$. +
Шаг 1. По условию $a_1 = 40$ т (за первый день), $n = 12$ дней, $S_{12} = 1008$ т (вся партия). Разность $d$ — неизвестна. +
Шаг 2. Подставим в формулу суммы: +$$S_{12} = \\dfrac{2\\cdot 40 + 11d}{2}\\cdot 12 = 6\\cdot(80 + 11d)$$ +По условию $S_{12} = 1008$: +$$6(80 + 11d) = 1008 \\implies 80 + 11d = 168 \\implies 11d = 88 \\implies d = 8$$ +Шаг 3. Находим объём за девятый день по формуле $a_n = a_1 + (n-1)d$: +$$a_9 = 40 + (9 - 1)\\cdot 8 = 40 + 64 = 104\\text{ т}$$ +
Ответ: $104$ т.
` + }, + { + text: `$ABCD$ — прямоугольник, точки $M$ и $K$ лежат на сторонах $AB$ и $CD$ соответственно, + $MK \\| AD$. Диагональ $BD$ пересекает отрезок $MK$ в точке $P$. + $S_{BMP} = 4$ см², $S_{PKD} = 9$ см². Найдите площадь прямоугольника $ABCD$.`, + sol: ` + + + + + + + + A + B + C + D + M + K + P + S = 4 + S = 9 + BM=p + AM=q + +Идея. $MK\\perp AB$ (т.к. $MK\\parallel AD$ и $AD\\perp AB$). Значит $\\triangle BMP$ — прямоугольный (∠M=90°), $\\triangle DKP$ — прямоугольный (∠K=90°). +
Шаг 1. Подобие. $\\triangle BMP \\sim \\triangle BAD$ (по двум углам: ∠B общий, $MP\\parallel AD$). Аналогично $\\triangle DKP\\sim\\triangle DCB$. +
Введём $BM=p$, $AM=q$. Так как $K$ под $M$: $DK=AM=q$. Высота прямоугольника $h=AD$. +
Из подобия: +$$MP = \\dfrac{BM}{BA}\\cdot AD = \\dfrac{p\\,h}{p+q}, \\quad PK = \\dfrac{DK}{DC}\\cdot CB = \\dfrac{q\\,h}{p+q}$$ +Шаг 2. Площади. +$$S_{BMP} = \\dfrac{1}{2}\\cdot p\\cdot\\dfrac{ph}{p+q} = \\dfrac{p^2 h}{2(p+q)} = 4$$ +$$S_{DKP} = \\dfrac{1}{2}\\cdot q\\cdot\\dfrac{qh}{p+q} = \\dfrac{q^2 h}{2(p+q)} = 9$$ +Шаг 3. Делим: $\\dfrac{p^2}{q^2}=\\dfrac{4}{9}$ → $\\dfrac{p}{q}=\\dfrac{2}{3}$. Пусть $p=2t$, $q=3t$. +
Из первого уравнения: +$$\\dfrac{4t^2 h}{2\\cdot5t} = \\dfrac{2th}{5}=4 \\implies th=10$$ +Шаг 4. Площадь прямоугольника. +$$S_{ABCD} = (p+q)\\cdot h = 5t\\cdot h = 5\\cdot10 = 50\\text{ см}^2$$ +Универсальная формула: $S_{ABCD} = 2\\bigl(\\sqrt{S_{BMP}}+\\sqrt{S_{DKP}}\\bigr)^{2} = 2(2+3)^2 = 50$. +
Ответ: $S_{ABCD} = 50$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v48.js b/frontend/js/exam9/variants/v48.js new file mode 100644 index 0000000..f531cc3 --- /dev/null +++ b/frontend/js/exam9/variants/v48.js @@ -0,0 +1,196 @@ +VARIANTS[48] = { + label: "Вариант 48", + tasks: [ + { + text: `Определите, какое из данных равенств является верным:`, + opts: [ + ["а", "$\\sqrt{50} = 2\\sqrt{5}$"], ["б", "$\\sqrt{50} = 5\\sqrt{2}$"], ["в", "$\\sqrt{50} = 25\\sqrt{2}$"], + ["г", "$\\sqrt{50} = 5\\sqrt{10}$"], ["д", "$\\sqrt{50} = 5$"], + ], + sol: `Разложим подкоренное число: $\\sqrt{50}=\\sqrt{25\\cdot2}=5\\sqrt{2}$. +
Ответ: б) $\\sqrt{50}=5\\sqrt{2}$
` + }, + { + text: `Значение выражения $\\dfrac{8^5}{8^3} + 8^1$ равно:`, + opts: [ + ["а", "$64$"], ["б", "$65$"], ["в", "$24$"], ["г", "$72$"], ["д", "$56$"], + ], + sol: `$\\dfrac{8^5}{8^3}+8^1=8^2+8=64+8=72$. +
Ответ: г) $72$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "диагонали любого ромба равны между собой;"], + ["б", "центр окружности, вписанной в треугольник, лежит на пересечении биссектрис треугольника;"], + ["в", "гипотенуза прямоугольного треугольника больше любого из катетов;"], + ["г", "угол, равный $91^{\\circ}$, — тупой?"], + ], + sol: `а) Диагонали ромба равны — НЕВЕРНО (равны только в квадрате). б)–г) верно. +
Ответ: а)
` + }, + { + text: `При каких значениях переменной $x$ равны значения трёхчленов + $12x^2 + 4 - 4x$ и $3 - 4x^2 + 4x$?`, + sol: `$12x^2+4-4x=3-4x^2+4x \\implies 16x^2-8x+1=0 \\implies (4x-1)^2=0 \\implies x=\\dfrac{1}{4}$. +
Ответ: $x=\\dfrac{1}{4}$
` + }, + { + text: `В прямоугольном треугольнике $ABC$ $\\angle A = 90^{\\circ}$, высота $AK = 24$ см, $BK = 18$ см. + Найдите косинус угла $C$.`, + sol: ` + + + + + + + + + B + C + A + K + 30 + 40 + BK=18 + KC=32 + 24 + +Свойство высоты прямоугольного треугольника: в прямоугольном треугольнике высота, проведённая из вершины прямого угла к гипотенузе, удовлетворяет: +
$h^2 = m\\cdot n$, где $m,\\,n$ — отрезки, на которые она делит гипотенузу. +
Шаг 1. В $\\triangle ABC$ прямой угол при $A$, высота $AK$ опущена на гипотенузу $BC$. Тогда: +$$AK^2 = BK\\cdot KC$$ +$$24^2 = 18\\cdot KC \\implies 576 = 18\\cdot KC \\implies KC = \\dfrac{576}{18} = 32\\text{ см}$$ +Шаг 2. Находим гипотенузу: +$$BC = BK + KC = 18 + 32 = 50\\text{ см}$$ +Шаг 3. Находим катет $AC$ по теореме Пифагора в $\\triangle AKC$: +$$AC^2 = AK^2 + KC^2 = 24^2 + 32^2 = 576 + 1024 = 1600$$ +$$AC = \\sqrt{1600} = 40\\text{ см}$$ +Шаг 4. По определению косинуса: $\\cos\\alpha = \\dfrac{\\text{прилежащий катет}}{\\text{гипотенуза}}$. +
В $\\triangle ABC$ для угла $C$: прилежащий катет — $AC$, гипотенуза — $BC$: +$$\\cos C = \\dfrac{AC}{BC} = \\dfrac{40}{50} = \\dfrac{4}{5} = 0{,}8$$ +
Ответ: $\\cos C=\\dfrac{4}{5}=0{,}8$
` + }, + { + text: `Упростите выражение + $\\dfrac{25 - 5n}{n^2-9} - \\dfrac{n}{n^2-9} : \\dfrac{n}{n+7} - \\dfrac{n-3}{n+3}$.`, + sol: `Порядок действий: сначала выполняется деление, потом сложение и вычитание. Формула разности квадратов: $a^2 - b^2 = (a-b)(a+b)$. +
Шаг 1. Раскладываем $n^2 - 9$ по формуле разности квадратов: +$$n^2 - 9 = (n - 3)(n + 3)$$ +ОДЗ: $n \\neq 3$, $n \\neq -3$, $n \\neq -7$, $n \\neq 0$. +
Шаг 2. Выполняем деление $\\dfrac{n}{n^2-9} : \\dfrac{n}{n+7}$ — умножаем на обратную дробь: +$$\\dfrac{n}{(n-3)(n+3)} \\cdot \\dfrac{n+7}{n} = \\dfrac{n+7}{(n-3)(n+3)}$$ +Шаг 3. Подставляем в выражение: +$$\\dfrac{25 - 5n}{(n-3)(n+3)} - \\dfrac{n+7}{(n-3)(n+3)} - \\dfrac{n-3}{n+3}$$ +Шаг 4. Объединяем первые две дроби с общим знаменателем: +$$\\dfrac{(25 - 5n) - (n + 7)}{(n-3)(n+3)} = \\dfrac{25 - 5n - n - 7}{(n-3)(n+3)} = \\dfrac{18 - 6n}{(n-3)(n+3)}$$ +В числителе выносим $-6$ за скобку: $18 - 6n = -6(n - 3)$. Тогда: +$$\\dfrac{-6(n-3)}{(n-3)(n+3)} = \\dfrac{-6}{n+3}$$ +Шаг 5. Остаётся вычесть третью дробь (общий знаменатель $n + 3$): +$$\\dfrac{-6}{n+3} - \\dfrac{n-3}{n+3} = \\dfrac{-6 - (n - 3)}{n+3} = \\dfrac{-6 - n + 3}{n+3} = \\dfrac{-n - 3}{n+3} = \\dfrac{-(n+3)}{n+3} = -1$$ +
Ответ: $-1$
` + }, + { + text: `График функции $f(x) = a(x-m)^2 + n$ изображён на рисунке. + Используя график функции, найдите $a$, $m$ и $n$. + Запишите формулу функции $y = f(x)$ в виде многочлена.`, + sol: `Вершинная форма квадратичной функции: $f(x) = a(x - m)^2 + n$ — это парабола с вершиной в точке $(m;\\,n)$. Знак $a$ определяет направление ветвей ($a \\gt 0$ — вверх, $a \\lt 0$ — вниз). +
Алгоритм нахождения $a$, $m$, $n$ по графику: +
    +
  • Шаг 1. Находим координаты вершины параболы — это $m$ (абсцисса) и $n$ (ордината).
    • +
  • Шаг 2. Берём любую другую точку $(x_0;\\,y_0)$ графика и подставляем в формулу: $a = \\dfrac{y_0 - n}{(x_0 - m)^2}$.
    • +
  • Шаг 3. Раскрываем скобки по формуле квадрата разности $(x - m)^2 = x^2 - 2mx + m^2$: +$$f(x) = a(x - m)^2 + n = ax^2 - 2am\\,x + (am^2 + n)$$
    • +
+Типичный пример (вершина $(2;\\,1)$, ветви вверх, график проходит через $(0;\\,5)$): +
— $m = 2$, $n = 1$. +
— Для $a$: $a = \\dfrac{5 - 1}{(0 - 2)^2} = \\dfrac{4}{4} = 1$. +
— Формула в виде многочлена: +$$f(x) = (x - 2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5$$ +
Ответ: читается с графика; $f(x) = ax^2 - 2am\\,x + (am^2 + n)$.
` + }, + { + text: `Найдите сумму целых решений системы неравенств + $$\\begin{cases} 9 - 2x < 0, \\\\[4pt] x^2 - 8x < -7. \\end{cases}$$`, + sol: `Метод решения системы неравенств: решаем каждое неравенство отдельно, затем находим пересечение решений (общую часть). +
Шаг 1. Решаем первое неравенство $9 - 2x \\lt 0$: +$$-2x \\lt -9$$ +Делим на $-2$ и меняем знак неравенства: +$$x \\gt 4{,}5$$ +Шаг 2. Решаем второе неравенство $x^2 - 8x \\lt -7$. Перенесём всё в одну сторону: +$$x^2 - 8x + 7 \\lt 0$$ +По теореме Виета разложим: ищем числа с суммой $8$ и произведением $7$ — это $1$ и $7$. +$$(x - 1)(x - 7) \\lt 0$$ +Парабола $y = x^2 - 8x + 7$ ветвями вверх; она отрицательна между корнями: +$$1 \\lt x \\lt 7$$ +Шаг 3. Пересечение условий $x \\gt 4{,}5$ и $1 \\lt x \\lt 7$: +$$4{,}5 \\lt x \\lt 7$$ +Шаг 4. Целые числа в этом промежутке: $x = 5,\\,6$. +$$\\text{Сумма} = 5 + 6 = 11$$ +
Ответ: $11$
` + }, + { + text: `Для перевозки партии щебня массой $880$ т фирма использует самосвал МАЗ-5551. + По плану норма перевозки ежедневно должна увеличиваться на одно и то же число тонн. + Известно, что за первый день было перевезено $30$ т щебня. + Определите, сколько тонн щебня было перевезено за шестой день, + если вся работа была выполнена за $11$ дней.`, + sol: `Метод арифметической прогрессии. По условию норма ежедневно увеличивается на одно и то же число тонн, значит дневные объёмы образуют арифметическую прогрессию. +
Формулы арифметической прогрессии: +
— $n$-й член: $a_n = a_1 + (n-1)d$. +
— Сумма первых $n$ членов: $S_n = \\dfrac{2a_1 + (n-1)d}{2}\\cdot n$. +
Шаг 1. По условию $a_1 = 30$ т, $n = 11$, $S_{11} = 880$ т. Разность $d$ — неизвестна. +
Шаг 2. Подставим в формулу суммы: +$$S_{11} = \\dfrac{2\\cdot 30 + 10d}{2}\\cdot 11 = (60 + 10d)\\cdot\\dfrac{11}{2} = 11\\cdot(30 + 5d)$$ +По условию $S_{11} = 880$: +$$11\\cdot(30 + 5d) = 880 \\implies 30 + 5d = 80 \\implies 5d = 50 \\implies d = 10$$ +Шаг 3. Находим объём за шестой день: +$$a_6 = a_1 + (6 - 1)\\cdot d = 30 + 5\\cdot 10 = 30 + 50 = 80\\text{ т}$$ +
Ответ: $80$ т
` + }, + { + text: `$ABCD$ — прямоугольник, точки $M$ и $K$ лежат на сторонах $AB$ и $CD$ соответственно, + $MK \\| AD$. Диагональ $AC$ пересекает отрезок $MK$ в точке $P$. + $S_{CPK} = 9$ см², $S_{AMP} = 16$ см². Найдите площадь прямоугольника $ABCD$.`, + sol: ` + + + + + + + + + + A + B + C + D + M + K + P + S = 16 + S = 9 + AM=p + MB=a−p + +Идея. Обозначим $AB = a$, $BC = h$ (стороны прямоугольника) и $AM = p$. Тогда $MB = a - p$. Так как $MK\\parallel AD$ ($AD$ — сторона, перпендикулярная $AB$), $MK = h$. +
Шаг 1. Диагональ $AC$ идёт от $A$ к $C$. По подобию треугольников ($AMP \\sim ABC$): $\\dfrac{AM}{AB} = \\dfrac{MP}{BC}$, откуда $MP = \\dfrac{p\\cdot h}{a}$. +
Тогда $PK = h - MP = \\dfrac{(a - p)\\cdot h}{a}$. +
Шаг 2. Площади прямоугольных треугольников $AMP$ и $CPK$: +$$S_{AMP} = \\dfrac{1}{2}\\cdot AM\\cdot MP = \\dfrac{1}{2}\\cdot p\\cdot\\dfrac{ph}{a} = \\dfrac{p^2 h}{2a} = 16$$ +$$S_{CPK} = \\dfrac{1}{2}\\cdot KC\\cdot PK = \\dfrac{1}{2}\\cdot(a - p)\\cdot\\dfrac{(a - p)h}{a} = \\dfrac{(a - p)^2 h}{2a} = 9$$ +Шаг 3. Делим равенства: +$$\\dfrac{S_{AMP}}{S_{CPK}} = \\dfrac{p^2}{(a - p)^2} = \\dfrac{16}{9}$$ +Извлекаем квадратный корень: +$$\\dfrac{p}{a - p} = \\dfrac{4}{3}$$ +Шаг 4. Параметризуем: пусть $p = 4t$, $a - p = 3t$, тогда $a = 7t$. +
Шаг 5. Подставим в $S_{AMP} = 16$: +$$\\dfrac{(4t)^2\\cdot h}{2\\cdot 7t} = 16 \\implies \\dfrac{16t^2 h}{14t} = 16 \\implies \\dfrac{8th}{7} = 16 \\implies th = 14$$ +Шаг 6. Площадь прямоугольника: +$$S_{ABCD} = a\\cdot h = 7t\\cdot h = 7\\cdot th = 7\\cdot 14 = 98\\text{ см}^2$$ +
Ответ: $98$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v49.js b/frontend/js/exam9/variants/v49.js new file mode 100644 index 0000000..572709c --- /dev/null +++ b/frontend/js/exam9/variants/v49.js @@ -0,0 +1,197 @@ +VARIANTS[49] = { + label: "Вариант 49", + tasks: [ + { + text: `Определите, какое из данных выражений НЕ является одночленом:`, + opts: [ + ["а", "$2abc$"], ["б", "$m^{11}$"], ["в", "$\\dfrac{4}{y}$"], + ["г", "$-\\dfrac{2}{7}c^4$"], ["д", "$\\dfrac{t}{5}$"], + ], + sol: `Определение: одночлен — произведение чисел и переменных в натуральных степенях.
+ Выражение $\\dfrac{4}{y}=4y^{-1}$ содержит переменную в знаменателе (отрицательная степень), + поэтому одночленом не является.
+ Остальные варианты — корректные одночлены. +
Ответ: в) $\\dfrac{4}{y}$.
` + }, + { + text: `Уравнение окружности с центром в точке $(5;\\; 0)$ и радиусом $\\sqrt{7}$ имеет вид:`, + opts: [ + ["а", "$(x+5)^2 + y^2 = 7$"], ["б", "$(x-5)^2 + y^2 = \\sqrt{7}$"], ["в", "$(x+5)^2 - y^2 = 7$"], + ["г", "$(x-5)^2 + y^2 = 7$"], ["д", "$(x-5)^2 - y^2 = 7$"], + ], + sol: `Уравнение окружности с центром $(a;\\,b)$ и радиусом $R$: + $$(x-a)^{2}+(y-b)^{2}=R^{2}.$$ + Подставляем $a=5,\\;b=0,\\;R=\\sqrt{7},\\;R^{2}=7$: + $$(x-5)^{2}+y^{2}=7.$$ +
Ответ: г) $(x-5)^{2}+y^{2}=7$.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "у подобных треугольников соответствующие углы равны;"], + ["б", "если прямые перпендикулярны, то угол между ними равен $90^{\\circ}$;"], + ["в", "$\\operatorname{tg} 45^{\\circ} = 1$;"], + ["г", "биссектриса любого треугольника делит сторону треугольника пополам?"], + ], + sol: `Проверим утверждения: +
    +
  • а) верно — определение подобных треугольников;
    • +
  • б) верно — определение перпендикулярных прямых;
    • +
  • в) верно — табличное значение $\\operatorname{tg}45^{\\circ}=1$;
    • +
  • г) неверно — пополам сторону делит медиана, а биссектриса + делит противоположную сторону в отношении прилежащих сторон.
    • +
+
Ответ: г).
` + }, + { + text: `Найдите значение выражения + $15^0 + \\sqrt{16} - \\left(\\dfrac{1}{3}\\right)^{-1} + \\sqrt{\\dfrac{1}{4}}$.`, + sol: `Вычисляем по частям: +
    +
  • $15^{0}=1;$
    • +
  • $\\sqrt{16}=4;$
    • +
  • $\\left(\\dfrac{1}{3}\\right)^{-1}=3;$
    • +
  • $\\sqrt{\\dfrac{1}{4}}=\\dfrac{1}{2}.$
    • +
+ Тогда $1+4-3+\\dfrac{1}{2}=2+\\dfrac{1}{2}=2{,}5.$ +
Ответ: $2{,}5$ (или $\\dfrac{5}{2}$).
` + }, + { + text: `$ABCD$ — параллелограмм, биссектриса угла $A$ пересекает сторону $BC$ в точке $K$, + $BK = 6$ см, $KC = 4$ см. Найдите периметр параллелограмма.`, + sol: ` + + + + + + + + A + B + C + D + K + 6 + 4 + + Свойство биссектрисы и параллельных прямых.
+ Шаг 1. Обозначим $\\angle A = 2\\alpha$. Так как $AK$ — биссектриса угла $A$, то она делит этот угол пополам, поэтому $\\angle BAK = \\alpha$.
+ Шаг 2. По свойству параллелограмма $BC \\parallel AD$. Прямая $AK$ — секущая для этих параллельных прямых, значит накрест лежащие углы равны: + $$\\angle AKB = \\angle KAD = \\alpha.$$ + Шаг 3. В $\\triangle ABK$ два угла равны ($\\angle BAK = \\angle AKB = \\alpha$), значит треугольник равнобедренный, и стороны напротив равных углов равны: + $$AB = BK = 6\\text{ см}.$$ + Шаг 4. Находим $BC$: точка $K$ лежит на стороне $BC$, поэтому + $$BC = BK + KC = 6 + 4 = 10\\text{ см}.$$ + Так как $AD = BC$ (противоположные стороны параллелограмма), то $AD = 10$ см.
+ Шаг 5. Периметр параллелограмма равен удвоенной сумме соседних сторон: + $$P = 2(AB + BC) = 2(6 + 10) = 32\\text{ см}.$$ +
Ответ: $32$ см.
` + }, + { + text: `Упростите выражение $\\dfrac{x^2 - 4x + 4}{(x+5)^2 - 49}$.`, + sol: `Формула квадрата разности: $a^2-2ab+b^2=(a-b)^2$.
+ Формула разности квадратов: $a^2-b^2=(a-b)(a+b)$.
+ Шаг 1. Раскладываем числитель по формуле квадрата разности (так как $4x=2\\cdot x\\cdot 2$ и $4=2^2$): + $$x^{2}-4x+4=(x-2)^{2}.$$ + Шаг 2. Раскладываем знаменатель по формуле разности квадратов (так как $49=7^2$): + $$(x+5)^{2}-49=(x+5-7)(x+5+7)=(x-2)(x+12).$$ + Шаг 3. Подставляем разложения в дробь и сокращаем общий множитель $(x-2)$: + $$\\dfrac{(x-2)^{2}}{(x-2)(x+12)}=\\dfrac{x-2}{x+12}.$$ + Шаг 4. Указываем ОДЗ: знаменатели исходного и сокращённого выражений не должны обращаться в ноль, поэтому $x\\ne 2$ и $x\\ne -12$. +
Ответ: $\\dfrac{x-2}{x+12}$.
` + }, + { + text: `Определите, сколько общих точек у прямой $y = -5$ и графика функции $y = -5x^2 - x + 1$. + В ответ запишите координаты точек пересечения.`, + sol: `Метод: чтобы найти общие точки графиков, приравнивают их правые части и решают полученное уравнение; каждый корень даёт одну общую точку.
+ Шаг 1. Приравниваем правые части (так как в общей точке значения $y$ совпадают): + $$-5=-5x^{2}-x+1.$$ + Шаг 2. Переносим всё в одну сторону: + $$5x^{2}+x-6=0.$$ + Шаг 3. Находим дискриминант квадратного уравнения по формуле $D=b^2-4ac$: + $$D=1^{2}-4\\cdot 5\\cdot(-6)=1+120=121,\\quad \\sqrt{D}=11.$$ + Шаг 4. По формуле корней $x_{1,2}=\\dfrac{-b\\pm\\sqrt{D}}{2a}$: + $$x_{1,2}=\\dfrac{-1\\pm 11}{10}\\;\\implies\\;x_{1}=1,\\;x_{2}=-\\dfrac{6}{5}=-1{,}2.$$ + Шаг 5. Так как уравнение имеет два различных корня, общих точек тоже две. При каждом из значений $y=-5$ (по условию прямой), значит точки пересечения: $(1;\\,-5)$ и $(-1{,}2;\\,-5)$. +
Ответ: $2$ точки: $(1;\\,-5)$ и $(-1{,}2;\\,-5)$.
` + }, + { + text: `Бригада маляров красит фасад здания площадью $2100$ м², ежедневно увеличивая норму покраски + на одно и то же число квадратных метров. Известно, что за первый и последний день + в сумме бригада покрасила $350$ м² фасада. + Определите, сколько дней бригада маляров красила весь фасад.`, + sol: `Формула суммы $n$ членов арифметической прогрессии: + $$S_n = \\dfrac{(a_1 + a_n) \\cdot n}{2}.$$ + Шаг 1. По условию ежедневные нормы покраски увеличиваются на одно и то же число, значит они образуют арифметическую прогрессию $a_1, a_2, \\ldots, a_n$, где $n$ — количество дней.
+ Шаг 2. По условию сумма площадей в первый и последний день: $a_1 + a_n = 350$ м². + Общая площадь — это сумма всех членов прогрессии: $S_n = 2100$ м².
+ Шаг 3. Подставляем в формулу суммы: + $$2100 = \\dfrac{350 \\cdot n}{2} = 175n.$$ + Шаг 4. Находим $n$: + $$n = \\dfrac{2100}{175} = 12.$$ +
Ответ: $12$ дней.
` + }, + { + text: `Найдите сумму целых решений системы неравенств + $$\\begin{cases} \\dfrac{x+5}{x} \\leq 0, \\\\[6pt] x^2 + 4x > -3. \\end{cases}$$`, + sol: `Метод интервалов: решаем каждое неравенство по отдельности, затем берём пересечение.
+ Шаг 1. Первое неравенство $\\dfrac{x+5}{x} \\leq 0$.
+ Находим нули числителя и знаменателя: $x = -5$ (включается, потому что $\\leq$) и $x = 0$ (выколота, так как делить на ноль нельзя).
+ Методом интервалов получаем $x \\in [-5;\\,0)$.
+ Шаг 2. Второе неравенство $x^2 + 4x \\gt -3$.
+ Переносим $-3$ влево: $x^2 + 4x + 3 \\gt 0$.
+ Раскладываем на множители: $x^2 + 4x + 3 = (x+1)(x+3)$.
+ Парабола ветвями вверх, поэтому $(x+1)(x+3) \\gt 0$ при $x \\lt -3$ или $x \\gt -1$, то есть + $$x \\in (-\\infty;\\,-3) \\cup (-1;\\,+\\infty).$$ + Шаг 3. Пересечение двух решений: + $$x \\in [-5;\\,-3) \\cup (-1;\\,0).$$ + Шаг 4. Целые решения.
+ На $[-5;\\,-3)$ — это $-5$ и $-4$ (число $-3$ не входит); на $(-1;\\,0)$ целых чисел нет.
+ Шаг 5. Сумма: $-5 + (-4) = -9$. +
Ответ: $-9$.
` + }, + { + text: `В треугольнике $ABC$ проведены отрезки $MK \\| AC$ и $KE \\| AB$, + где точки $M$, $K$ и $E$ принадлежат сторонам $AB$, $BC$ и $AC$ соответственно. + Площадь треугольника $MBK$ равна $9$ см², треугольника $EKC$ — $16$ см². + Найдите площадь четырёхугольника $AMKE$.`, + sol: ` + + + + + + A + B + C + M + K + E + S=9 + S=16 + AMKE = ? + + Теорема: отношение площадей подобных треугольников равно квадрату коэффициента подобия.
+ Шаг 1. Подобие $\\triangle MBK \\sim \\triangle ABC$.
+ Так как $MK \\parallel AC$, то $\\triangle MBK$ подобен $\\triangle ABC$ (углы $B$ общий, остальные — как соответственные при параллельных). Коэффициент подобия: + $$k_1 = \\dfrac{BK}{BC}.$$ + Шаг 2. Подобие $\\triangle KEC \\sim \\triangle ABC$.
+ Так как $KE \\parallel AB$, аналогично $\\triangle KEC \\sim \\triangle ABC$ с коэффициентом + $$k_2 = \\dfrac{KC}{BC}.$$ + Шаг 3. Связь коэффициентов.
+ Так как $BK + KC = BC$, то + $$k_1 + k_2 = \\dfrac{BK + KC}{BC} = 1.$$ + Шаг 4. Выражаем коэффициенты через площади.
+ Пусть $S = S_{ABC}$. По теореме об отношении площадей: + $$\\dfrac{S_{MBK}}{S} = k_1^2 \\implies k_1 = \\dfrac{3}{\\sqrt{S}}; \\quad \\dfrac{S_{EKC}}{S} = k_2^2 \\implies k_2 = \\dfrac{4}{\\sqrt{S}}.$$ + Шаг 5. Находим $S$.
+ Подставляем в равенство $k_1 + k_2 = 1$: + $$\\dfrac{3}{\\sqrt{S}} + \\dfrac{4}{\\sqrt{S}} = 1 \\implies \\dfrac{7}{\\sqrt{S}} = 1 \\implies \\sqrt{S} = 7 \\implies S = 49.$$ + Шаг 6. Площадь четырёхугольника $AMKE$.
+ Большой треугольник разбит на два маленьких и четырёхугольник: + $$S_{AMKE} = S_{ABC} - S_{MBK} - S_{EKC} = 49 - 9 - 16 = 24\\text{ см}^2.$$ +
Ответ: $24$ см².
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v50.js b/frontend/js/exam9/variants/v50.js new file mode 100644 index 0000000..3cb6b99 --- /dev/null +++ b/frontend/js/exam9/variants/v50.js @@ -0,0 +1,187 @@ +VARIANTS[50] = { + label: "Вариант 50", + tasks: [ + { + text: `Определите, какое из данных выражений НЕ является одночленом:`, + opts: [ + ["а", "$n^{12}$"], ["б", "$-\\dfrac{3}{8}b^3$"], ["в", "$\\dfrac{5}{z}$"], + ["г", "$3abc$"], ["д", "$1$"], + ], + sol: `Определение: одночлен — произведение чисел и переменных в натуральных степенях.
+ Выражение $\\dfrac{5}{z}=5z^{-1}$ содержит переменную в знаменателе (отрицательная степень), + поэтому одночленом не является.
+ Остальные варианты — корректные одночлены. +
Ответ: в) $\\dfrac{5}{z}$.
` + }, + { + text: `Уравнение окружности с центром в точке $(0;\\; 4)$ и радиусом $\\sqrt{5}$ имеет вид:`, + opts: [ + ["а", "$x^2 + (y+4)^2 = 5$"], ["б", "$x^2 + (y-4)^2 = 5$"], ["в", "$x^2 - (y+4)^2 = 5$"], + ["г", "$x^2 - (y-4)^2 = 5$"], ["д", "$x^2 + (y-4)^2 = \\sqrt{5}$"], + ], + sol: `Уравнение окружности с центром $(a;\\,b)$ и радиусом $R$: + $$(x-a)^{2}+(y-b)^{2}=R^{2}.$$ + Подставляем $a=0,\\;b=4,\\;R=\\sqrt{5},\\;R^{2}=5$: + $$x^{2}+(y-4)^{2}=5.$$ +
Ответ: б) $x^{2}+(y-4)^{2}=5$.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "у подобных треугольников соответствующие стороны пропорциональны;"], + ["б", "$\\operatorname{ctg} 45^{\\circ} = 1$;"], + ["в", "если угол между прямыми равен $90^{\\circ}$, то они перпендикулярны;"], + ["г", "медиана любого треугольника перпендикулярна стороне, к которой проведена?"], + ], + sol: `Проверим утверждения: +
    +
  • а) верно — определение подобных треугольников;
    • +
  • б) верно — табличное значение $\\operatorname{ctg}45^{\\circ}=1$;
    • +
  • в) верно — определение перпендикулярных прямых;
    • +
  • г) неверно — медиана соединяет вершину с серединой противоположной стороны, + но в общем случае она не перпендикулярна этой стороне.
    • +
+
Ответ: г).
` + }, + { + text: `Найдите значение выражения + $12^0 + \\sqrt{36} - \\left(\\dfrac{1}{2}\\right)^{-1} + \\sqrt{\\dfrac{1}{16}}$.`, + sol: `Вычисляем по частям: +
    +
  • $12^{0}=1;$
    • +
  • $\\sqrt{36}=6;$
    • +
  • $\\left(\\dfrac{1}{2}\\right)^{-1}=2;$
    • +
  • $\\sqrt{\\dfrac{1}{16}}=\\dfrac{1}{4}.$
    • +
+ Тогда $1+6-2+\\dfrac{1}{4}=5+\\dfrac{1}{4}=5{,}25.$ +
Ответ: $5{,}25$ (или $\\dfrac{21}{4}$).
` + }, + { + text: `$ABCD$ — параллелограмм, $DC = 12$ см. Биссектриса угла $B$ пересекает сторону $AD$ + в точке $M$, $MD = 4$ см. Найдите периметр параллелограмма.`, + sol: ` + + + + A + B + C + D + M + AM=12 + MD=4 + + Свойство биссектрисы и параллельных прямых.
+ Шаг 1. Пусть $\\angle B = 2\\beta$. Так как $BM$ — биссектриса угла $B$, то $\\angle ABM = \\angle MBC = \\beta$.
+ Шаг 2. По свойству параллелограмма $AD \\parallel BC$. $BM$ — секущая, значит накрест лежащие углы равны: + $$\\angle BMA = \\angle MBC = \\beta.$$ + Шаг 3. В $\\triangle ABM$ два угла равны ($\\angle ABM = \\angle BMA = \\beta$), значит он равнобедренный, и стороны напротив равных углов равны: + $$AB = AM.$$ + Шаг 4. Так как в параллелограмме противоположные стороны равны, $AB = DC = 12$ см. Значит $AM = 12$ см.
+ Шаг 5. Находим $AD$: точка $M$ лежит на стороне $AD$, поэтому + $$AD = AM + MD = 12 + 4 = 16\\text{ см}.$$ + А $BC = AD = 16$ см (противоположные стороны параллелограмма).
+ Шаг 6. Периметр параллелограмма: + $$P = 2(AB + BC) = 2(12 + 16) = 56\\text{ см}.$$ +
Ответ: $56$ см.
` + }, + { + text: `Упростите выражение $\\dfrac{y^2 + 14y + 49}{(y+3)^2 - 16}$.`, + sol: `Формула квадрата суммы: $a^2+2ab+b^2=(a+b)^2$.
+ Формула разности квадратов: $a^2-b^2=(a-b)(a+b)$.
+ Шаг 1. Раскладываем числитель по формуле квадрата суммы (так как $14y=2\\cdot y\\cdot 7$ и $49=7^2$): + $$y^{2}+14y+49=(y+7)^{2}.$$ + Шаг 2. Раскладываем знаменатель по формуле разности квадратов (так как $16=4^2$): + $$(y+3)^{2}-16=(y+3-4)(y+3+4)=(y-1)(y+7).$$ + Шаг 3. Подставляем разложения и сокращаем общий множитель $(y+7)$: + $$\\dfrac{(y+7)^{2}}{(y-1)(y+7)}=\\dfrac{y+7}{y-1}.$$ + Шаг 4. ОДЗ: знаменатели исходного и сокращённого выражений не должны быть равны нулю, значит $y\\ne 1$ и $y\\ne -7$. +
Ответ: $\\dfrac{y+7}{y-1}$.
` + }, + { + text: `Определите, сколько общих точек у прямой $y = -6$ и графика функции $y = -4x^2 + x - 1$. + В ответ запишите координаты точек пересечения.`, + sol: `Метод: в общей точке двух графиков ординаты совпадают. Поэтому приравниваем правые части и считаем количество корней.
+ Шаг 1. Приравниваем правые части уравнений: + $$-6=-4x^{2}+x-1.$$ + Шаг 2. Переносим в одну часть и приводим к стандартному виду: + $$4x^{2}-x-5=0.$$ + Шаг 3. Считаем дискриминант по формуле $D=b^2-4ac$: + $$D=(-1)^{2}-4\\cdot 4\\cdot(-5)=1+80=81,\\quad \\sqrt{D}=9.$$ + Шаг 4. Находим корни по формуле $x_{1,2}=\\dfrac{-b\\pm\\sqrt{D}}{2a}$: + $$x_{1,2}=\\dfrac{1\\pm 9}{8}\\;\\implies\\;x_{1}=\\dfrac{5}{4},\\;x_{2}=-1.$$ + Шаг 5. Уравнение имеет два корня, значит общих точек две. При обоих значениях $y=-6$ (так как точки лежат на прямой $y=-6$). +
Ответ: $2$ точки: $\\left(\\dfrac{5}{4};\\,-6\\right)$ и $(-1;\\,-6)$.
` + }, + { + text: `Бригада маляров красит фасад здания площадью $2700$ м², ежедневно увеличивая норму покраски + на одно и то же число квадратных метров. Известно, что за первый и последний день + в сумме бригада покрасила $360$ м² фасада. + Определите, сколько дней бригада маляров красила весь фасад.`, + sol: `Формула суммы $n$ членов арифметической прогрессии: + $$S_n = \\dfrac{(a_1 + a_n) \\cdot n}{2}.$$ + Шаг 1. Так как ежедневные нормы увеличиваются на одно и то же число, они образуют арифметическую прогрессию $a_1, a_2, \\ldots, a_n$, где $n$ — искомое число дней.
+ Шаг 2. По условию $a_1 + a_n = 360$ м² и общая площадь $S_n = 2700$ м².
+ Шаг 3. Подставляем в формулу суммы: + $$2700 = \\dfrac{360 \\cdot n}{2} = 180n.$$ + Шаг 4. Находим $n$: + $$n = \\dfrac{2700}{180} = 15.$$ +
Ответ: $15$ дней.
` + }, + { + text: `Найдите сумму целых решений системы неравенств + $$\\begin{cases} \\dfrac{x-3}{x+5} \\leq 0, \\\\[6pt] x^2 + 3x > -2. \\end{cases}$$`, + sol: `Метод интервалов: решаем каждое неравенство и находим пересечение.
+ Шаг 1. Первое неравенство $\\dfrac{x-3}{x+5} \\leq 0$.
+ Нули числителя и знаменателя: $x = 3$ (входит, потому что $\\leq$) и $x = -5$ (выколота, нельзя делить на ноль).
+ Методом интервалов: $x \\in (-5;\\,3]$.
+ Шаг 2. Второе неравенство $x^2 + 3x \\gt -2$.
+ Переносим: $x^2 + 3x + 2 \\gt 0$, раскладываем: $(x+1)(x+2) \\gt 0$.
+ Парабола ветвями вверх, поэтому + $$x \\in (-\\infty;\\,-2) \\cup (-1;\\,+\\infty).$$ + Шаг 3. Пересечение: + $$x \\in (-5;\\,-2) \\cup (-1;\\,3].$$ + Шаг 4. Целые решения.
+ На $(-5;\\,-2)$ — это $-4$ и $-3$; на $(-1;\\,3]$ — это $0, 1, 2, 3$.
+ Шаг 5. Сумма: $(-4) + (-3) + 0 + 1 + 2 + 3 = -1$. +
Ответ: $-1$.
` + }, + { + text: `В треугольнике $ABC$ проведены отрезки $MK \\| AC$ и $KE \\| AB$, + где точки $M$, $K$ и $E$ принадлежат сторонам $AB$, $BC$ и $AC$ соответственно. + Площадь треугольника $MBK$ равна $16$ см², треугольника $EKC$ — $25$ см². + Найдите площадь четырёхугольника $AMKE$.`, + sol: ` + + + + + A + B + C + M + K + E + S=16 + S=25 + AMKE = ? + + Теорема: отношение площадей подобных треугольников равно квадрату коэффициента подобия.
+ Шаг 1. Подобие $\\triangle MBK \\sim \\triangle ABC$.
+ Так как $MK \\parallel AC$, треугольники подобны с коэффициентом + $$k_1 = \\dfrac{BK}{BC}.$$ + Шаг 2. Подобие $\\triangle KEC \\sim \\triangle ABC$.
+ Так как $KE \\parallel AB$, аналогично + $$k_2 = \\dfrac{KC}{BC}.$$ + Шаг 3. Так как $BK + KC = BC$, то $k_1 + k_2 = 1$.
+ Шаг 4. Выражаем коэффициенты через площади.
+ Пусть $S = S_{ABC}$. Тогда + $$\\dfrac{S_{MBK}}{S} = k_1^2 \\implies k_1 = \\dfrac{4}{\\sqrt{S}}; \\quad \\dfrac{S_{EKC}}{S} = k_2^2 \\implies k_2 = \\dfrac{5}{\\sqrt{S}}.$$ + Шаг 5. Находим $S$. + $$\\dfrac{4}{\\sqrt{S}} + \\dfrac{5}{\\sqrt{S}} = 1 \\implies \\dfrac{9}{\\sqrt{S}} = 1 \\implies \\sqrt{S} = 9 \\implies S = 81.$$ + Шаг 6. Площадь четырёхугольника $AMKE$. + $$S_{AMKE} = S - S_{MBK} - S_{EKC} = 81 - 16 - 25 = 40\\text{ см}^2.$$ +
Ответ: $40$ см².
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v51.js b/frontend/js/exam9/variants/v51.js new file mode 100644 index 0000000..e8493ad --- /dev/null +++ b/frontend/js/exam9/variants/v51.js @@ -0,0 +1,203 @@ +VARIANTS[51] = { + label: "Вариант 51", + tasks: [ + { + text: `Сумма корней квадратного уравнения $x^2 + 6x - 11 = 0$ равна:`, + opts: [ + ["а", "$6$"], ["б", "$-11$"], ["в", "$-6$"], ["г", "$11$"], ["д", "$4$"], + ], + sol: `По теореме Виета для уравнения $x^2 + px + q = 0$: + сумма корней равна $-p$.
+ Здесь $p = 6$, поэтому $x_1 + x_2 = -6$. +
Ответ: в) $-6$.
` + }, + { + text: `Запись выражения $5^2 : 25 \\cdot 5^5$ в виде степени с основанием $5$ имеет вид:`, + opts: [ + ["а", "$5^7$"], ["б", "$5^8$"], ["в", "$5^9$"], ["г", "$5^5$"], ["д", "$5^1$"], + ], + sol: `Заменим $25 = 5^2$:
+ $5^2 : 25 \\cdot 5^5 = \\dfrac{5^2}{5^2} \\cdot 5^5 = 1 \\cdot 5^5 = 5^5.$ +
Ответ: г) $5^5$.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "площадь прямоугольника равна произведению его соседних сторон;"], + ["б", "диагонали любого ромба равны между собой;"], + ["в", "диаметры одной окружности равны между собой;"], + ["г", "если три угла одного треугольника равны трём углам другого треугольника, то треугольники подобны?"], + ], + sol: `Проверим утверждения: +
    +
  • а) верно — формула площади прямоугольника;
    • +
  • б) НЕ верно — диагонали ромба в общем случае не равны; + они равны только в частном случае — квадрате;
    • +
  • в) верно — все диаметры окружности равны $2R$;
    • +
  • г) верно — признак подобия треугольников по трём углам.
    • +
+
Ответ: б).
` + }, + { + text: `Найдите частное от деления наименьшего общего кратного на наибольший общий делитель + чисел $112$ и $80$.`, + sol: `Разложим числа на простые множители:
+ $112 = 2^4 \\cdot 7,\\quad 80 = 2^4 \\cdot 5.$
+ НОД$(112,\\,80) = 2^4 = 16.$
+ НОК$(112,\\,80) = 2^4 \\cdot 5 \\cdot 7 = 560.$
+ $\\dfrac{\\text{НОК}}{\\text{НОД}} = \\dfrac{560}{16} = 35.$ +
Ответ: $35$.
` + }, + { + text: `Сократите дробь $\\dfrac{b + 2\\sqrt{b} + 1}{\\sqrt{b} + b}$.`, + sol: `Формула квадрата суммы: $(a + c)^2 = a^2 + 2ac + c^2$.
+ Шаг 1. Преобразуем числитель.
+ Заметим, что $b = (\\sqrt{b})^2$ и $1 = 1^2$. Тогда $b + 2\\sqrt{b} + 1$ — это полный квадрат: + $$b + 2\\sqrt{b} + 1 = (\\sqrt{b} + 1)^2.$$ + Шаг 2. Преобразуем знаменатель.
+ Вынесем общий множитель $\\sqrt{b}$: + $$\\sqrt{b} + b = \\sqrt{b} + \\sqrt{b} \\cdot \\sqrt{b} = \\sqrt{b}(1 + \\sqrt{b}).$$ + Шаг 3. Сокращаем дробь.
+ В числителе и знаменателе есть общий множитель $(\\sqrt{b} + 1)$: + $$\\dfrac{(\\sqrt{b} + 1)^2}{\\sqrt{b}(\\sqrt{b} + 1)} = \\dfrac{\\sqrt{b} + 1}{\\sqrt{b}}.$$ +
Ответ: $\\dfrac{\\sqrt{b}+1}{\\sqrt{b}}$ (или $1 + \\dfrac{1}{\\sqrt{b}}$).
` + }, + { + text: `Высота $DH$ ромба $ABCD$ делит сторону $BC$ на отрезки $BH = 8$ см и $HC = 12$ см. + Найдите площадь ромба.`, + sol: ` + + + + A + B + C + D + H + BH=8 + HC=12 + DH +
+ Свойство ромба: все стороны равны.
+ Теорема Пифагора: в прямоугольном треугольнике $c^2 = a^2 + b^2$, где $c$ — гипотенуза.
+ Формула площади ромба через высоту: $S = a \\cdot h$, где $a$ — сторона, $h$ — высота.
+ Шаг 1. Находим сторону ромба.
+ Точка $H$ лежит на стороне $BC$, поэтому + $$BC = BH + HC = 8 + 12 = 20\\text{ см}.$$ + Так как все стороны ромба равны, $DC = BC = 20$ см.
+ Шаг 2. Находим высоту $DH$ по теореме Пифагора.
+ $DH$ — высота, значит $\\triangle DHC$ прямоугольный с прямым углом в $H$. Здесь $DC = 20$ — гипотенуза, $HC = 12$ — катет: + $$DH = \\sqrt{DC^2 - HC^2} = \\sqrt{400 - 144} = \\sqrt{256} = 16\\text{ см}.$$ + Шаг 3. Находим площадь. + $$S = BC \\cdot DH = 20 \\cdot 16 = 320\\text{ см}^2.$$ +
Ответ: $320$ см$^2$.
` + }, + { + text: `Найдите наименьшее целое значение аргумента, принадлежащее области определения функции + $y = \\dfrac{\\sqrt{x+8}}{x^2 - 2x - 80}$.`, + sol: `Правила нахождения ОДЗ: +
    +
  • выражение под корнем должно быть $\\geq 0$;
    • +
  • знаменатель не должен равняться нулю.
    • +
+ Шаг 1. Условие подкоренного выражения.
+ Под корнем стоит $x + 8$, значит + $$x + 8 \\geq 0 \\implies x \\geq -8.$$ + Шаг 2. Условие знаменателя.
+ Знаменатель $x^2 - 2x - 80$ не должен быть равен нулю. Разложим его на множители (находим корни, например по теореме Виета: $-10 \\cdot 8 = -80$, $-10 + 8 \\neq 2$; попробуем $10$ и $-8$: $10 \\cdot (-8) = -80$, $10 + (-8) = 2$ — подходит): + $$x^2 - 2x - 80 = (x - 10)(x + 8) \\neq 0 \\implies x \\neq 10\\text{ и }x \\neq -8.$$ + Шаг 3. Объединяем условия.
+ $x \\geq -8$ и одновременно $x \\neq -8$, $x \\neq 10$. Значит $x \\gt -8,\\; x \\neq 10$.
+ Шаг 4. Наименьшее целое.
+ Наименьшее целое число, большее $-8$, — это $-7$. +
Ответ: $-7$.
` + }, + { + text: `Семья Ивановых ежемесячно вносит плату за коммунальные услуги, телефон и электричество. + Если бы коммунальные услуги подорожали на $50\\%$, общая сумма платежа увеличилась бы на $25\\%$. + Если бы электричество подорожало на $50\\%$, общая сумма платежа увеличилась бы на $20\\%$. + Какой процент от общей суммы платежа приходится на телефон?`, + sol: `Метод введения переменных и составления уравнений по условию задачи.
+ Шаг 1. Вводим переменные: пусть $У$ — плата за коммунальные услуги, $Т$ — за телефон, $Э$ — за электричество. Тогда общая сумма платежа + $$S = У + Т + Э.$$ + Шаг 2. Используем первое условие. Если коммунальные услуги подорожают на $50\\%$, то прибавка к их стоимости составит $0{,}5\\,У$. По условию эта же прибавка равна $25\\%$ от общей суммы, то есть $0{,}25\\,S$. Значит: + $$0{,}5\\,У = 0{,}25\\,S \\implies У = 0{,}5\\,S,$$ + то есть на услуги приходится $50\\%$ общей суммы.
+ Шаг 3. Используем второе условие. Подорожание электричества на $50\\%$ — это прибавка $0{,}5\\,Э$, и она равна $20\\%$ общей суммы: + $$0{,}5\\,Э = 0{,}2\\,S \\implies Э = 0{,}4\\,S,$$ + то есть на электричество приходится $40\\%$ суммы.
+ Шаг 4. Находим долю телефона. Так как $S = У + Т + Э$, то + $$Т = S - У - Э = S - 0{,}5\\,S - 0{,}4\\,S = 0{,}1\\,S = 10\\%\\,S.$$ +
Ответ: $10\\%$.
` + }, + { + text: `Определите количество целых решений неравенства + $\\dfrac{3x^2 + 10x + 3}{(3-x)^2(4-x^2)} > 0$.`, + sol: `Числитель: $3x^2 + 10x + 3 = (3x+1)(x+3),$ + корни $x = -3,\\; x = -\\tfrac{1}{3}.$
+ Знаменатель: $(3-x)^2(4-x^2) = (3-x)^2(2-x)(2+x).$
+ Множитель $(3-x)^2 \\ge 0,$ обращается в $0$ при $x = 3$ (исключается из ОДЗ); + на знак не влияет. Корни знаменателя: $x = -2,\\; x = 2,\\; x = 3.$
+ Метод интервалов (критические точки $-3,\\,-2,\\,-\\tfrac{1}{3},\\,2,\\,3$): + + + + + + + + + +
ИнтервалЗнак
$x<-3$$-$
$(-3;-2)$$+$
$(-2;-\\tfrac{1}{3})$$-$
$(-\\tfrac{1}{3};\\,2)$$+$
$(2;\\,3)$$-$
$x>3$$-$
+ Решение: $x \\in (-3;\\,-2) \\cup \\left(-\\tfrac{1}{3};\\,2\\right).$
+ Целые в этих интервалах: в $(-3;-2)$ — нет; в $(-\\tfrac{1}{3};\\,2)$ — это $0$ и $1.$ + Всего $2$ целых решения. +
Ответ: $2$.
` + }, + { + text: `Найдите площадь прямоугольной трапеции с основаниями $4$ см и $12$ см, + если известно, что в трапецию можно вписать окружность.`, + sol: ` + + + + + + + + + + + + + A + B + C + D + + 4 + 12 + h=6 + 10 + r=3 + + Свойство описанного четырёхугольника (с вписанной окружностью): + суммы противоположных сторон равны.
+ Значит, сумма оснований равна сумме боковых сторон: + $4 + 12 = a + b \\implies a + b = 16,$ + где $a$ — боковая сторона, перпендикулярная основаниям (равна высоте $h$), + а $b$ — наклонная боковая.
+ Опустив высоту из вершины меньшего основания, получим прямоугольный треугольник + с катетами $h$ и $12 - 4 = 8$ и гипотенузой $b$:
+ $b = \\sqrt{h^2 + 8^2} = \\sqrt{h^2 + 64}.$
+ Подставим $a = h$: + $h + \\sqrt{h^2 + 64} = 16 \\implies \\sqrt{h^2 + 64} = 16 - h$
+ $\\implies h^2 + 64 = 256 - 32h + h^2 \\implies 32h = 192 \\implies h = 6$ см.
+ Площадь трапеции: + $S = \\dfrac{4 + 12}{2} \\cdot 6 = 8 \\cdot 6 = 48$ см$^2.$ +
Ответ: $48$ см$^2$.
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v52.js b/frontend/js/exam9/variants/v52.js new file mode 100644 index 0000000..b84e6e5 --- /dev/null +++ b/frontend/js/exam9/variants/v52.js @@ -0,0 +1,200 @@ +VARIANTS[52] = { + label: "Вариант 52", + tasks: [ + { + text: `Сумма корней квадратного уравнения $x^2 + 7x - 13 = 0$ равна:`, + opts: [ + ["а", "$-13$"], ["б", "$-7$"], ["в", "$7$"], ["г", "$13$"], ["д", "$20$"], + ], + sol: `По теореме Виета для уравнения $x^2 + px + q = 0$: + сумма корней равна $-p$.
+ Здесь $p = 7$, поэтому $x_1 + x_2 = -7$. +
Ответ: б) $-7$.
` + }, + { + text: `Запись выражения $27 : 3^3 \\cdot 3^7$ в виде степени с основанием $3$ имеет вид:`, + opts: [ + ["а", "$3^9$"], ["б", "$3^8$"], ["в", "$3^7$"], ["г", "$3^6$"], ["д", "$3^1$"], + ], + sol: `Заменим $27 = 3^3$:
+ $27 : 3^3 \\cdot 3^7 = \\dfrac{3^3}{3^3} \\cdot 3^7 = 1 \\cdot 3^7 = 3^7.$ +
Ответ: в) $3^7$.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "диагонали любого прямоугольника перпендикулярны;"], + ["б", "площадь квадрата равна квадрату его стороны;"], + ["в", "радиусы одной окружности равны между собой;"], + ["г", "если три стороны одного треугольника соответственно равны трём сторонам другого, то треугольники равны?"], + ], + sol: `Проверим утверждения: +
    +
  • а) НЕ верно — диагонали прямоугольника в общем случае не перпендикулярны; + они перпендикулярны только в квадрате;
    • +
  • б) верно — формула площади квадрата $S = a^2$;
    • +
  • в) верно — все радиусы окружности равны $R$;
    • +
  • г) верно — признак равенства треугольников «три стороны».
    • +
+
Ответ: а).
` + }, + { + text: `Найдите частное от деления наименьшего общего кратного на наибольший общий делитель + чисел $64$ и $288$.`, + sol: `Разложим числа на простые множители:
+ $64 = 2^{6},\\quad 288 = 2^{5}\\cdot 3^{2}.$
+ НОД$(64,\\,288) = 2^{5} = 32.$
+ НОК$(64,\\,288) = 2^{6}\\cdot 3^{2} = 576.$
+ $\\dfrac{\\text{НОК}}{\\text{НОД}} = \\dfrac{576}{32} = 18.$ +
Ответ: $18$.
` + }, + { + text: `Сократите дробь $\\dfrac{a - 9}{a - 6\\sqrt{a} + 9}$.`, + sol: `Формула разности квадратов: $x^2 - y^2 = (x - y)(x + y)$.
+ Формула квадрата разности: $(x - y)^2 = x^2 - 2xy + y^2$.
+ Шаг 1. Преобразуем числитель.
+ Запишем $a = (\\sqrt{a})^2$ и $9 = 3^2$, тогда числитель — разность квадратов: + $$a - 9 = (\\sqrt{a})^2 - 3^2 = (\\sqrt{a} - 3)(\\sqrt{a} + 3).$$ + Шаг 2. Преобразуем знаменатель.
+ $a - 6\\sqrt{a} + 9 = (\\sqrt{a})^2 - 2 \\cdot \\sqrt{a} \\cdot 3 + 3^2$ — полный квадрат: + $$a - 6\\sqrt{a} + 9 = (\\sqrt{a} - 3)^2.$$ + Шаг 3. Сокращаем дробь.
+ Общий множитель — $(\\sqrt{a} - 3)$: + $$\\dfrac{(\\sqrt{a} - 3)(\\sqrt{a} + 3)}{(\\sqrt{a} - 3)^2} = \\dfrac{\\sqrt{a} + 3}{\\sqrt{a} - 3},\\quad a \\geq 0,\\; a \\neq 9.$$ +
Ответ: $\\dfrac{\\sqrt{a}+3}{\\sqrt{a}-3}$.
` + }, + { + text: `Высота $BH$ ромба $ABCD$ делит сторону $AD$ на отрезки $AH = 5$ см и $HD = 8$ см. + Найдите площадь ромба.`, + sol: ` + + + + + A + B + C + D + H + AH=5 + HD=8 + BH +
+ Свойство ромба: все стороны равны.
+ Теорема Пифагора: $c^2 = a^2 + b^2$, где $c$ — гипотенуза.
+ Формула площади ромба через высоту: $S = a \\cdot h$.
+ Шаг 1. Находим сторону ромба.
+ Точка $H$ лежит на $AD$, поэтому + $$AD = AH + HD = 5 + 8 = 13\\text{ см}.$$ + Все стороны ромба равны, значит $AB = AD = 13$ см.
+ Шаг 2. Находим высоту $BH$ по теореме Пифагора.
+ $BH$ — высота, $\\triangle ABH$ прямоугольный с прямым углом в $H$. Гипотенуза $AB = 13$, катет $AH = 5$: + $$BH = \\sqrt{AB^2 - AH^2} = \\sqrt{169 - 25} = \\sqrt{144} = 12\\text{ см}.$$ + Шаг 3. Находим площадь. + $$S = AD \\cdot BH = 13 \\cdot 12 = 156\\text{ см}^2.$$ +
Ответ: $156$ см$^2$.
` + }, + { + text: `Найдите наименьшее целое значение аргумента, принадлежащее области определения функции + $y = \\dfrac{\\sqrt{x+12}}{x^2 - 2x - 120}$.`, + sol: `Правила нахождения ОДЗ: +
    +
  • выражение под квадратным корнем должно быть $\\geq 0$;
    • +
  • знаменатель не может быть равен нулю.
    • +
+ Шаг 1. Условие подкоренного выражения.
+ $$x + 12 \\geq 0 \\implies x \\geq -12.$$ + Шаг 2. Условие знаменателя.
+ Раскладываем $x^2 - 2x - 120$ на множители (по теореме Виета подбираем числа $-12$ и $10$: $-12 \\cdot 10 = -120$, $-12 + 10 = -2$): + $$x^2 - 2x - 120 = (x - 12)(x + 10) \\neq 0 \\implies x \\neq 12,\\; x \\neq -10.$$ + Шаг 3. Объединяем условия.
+ $x \\geq -12$, $x \\neq -10$, $x \\neq 12$.
+ Шаг 4. Проверяем $x = -12$.
+ Подкоренное: $-12 + 12 = 0 \\geq 0$ — допустимо.
+ Знаменатель: $144 - 2 \\cdot (-12) - 120 = 144 + 24 - 120 = 48 \\neq 0$ — допустимо.
+ Шаг 5. Значит наименьшее целое значение $x = -12$. +
Ответ: $-12$.
` + }, + { + text: `Семья Петровых ежемесячно вносит плату за коммунальные услуги, телефон и электричество. + Если бы коммунальные услуги подорожали на $50\\%$, общая сумма платежа увеличилась бы на $35\\%$. + Если бы электричество подорожало на $50\\%$, общая сумма платежа увеличилась бы на $10\\%$. + Какой процент от общей суммы платежа приходится на телефон?`, + sol: `Метод введения переменных и составления уравнений по условию задачи.
+ Шаг 1. Вводим переменные: пусть $У$ — плата за коммунальные услуги, $Т$ — за телефон, $Э$ — за электричество. Общая сумма платежа + $$S = У + Т + Э.$$ + Шаг 2. Используем первое условие. Подорожание услуг на $50\\%$ — это прибавка $0{,}5\\,У$, и она равна $35\\%$ от общей суммы: + $$0{,}5\\,У = 0{,}35\\,S \\implies У = 0{,}7\\,S,$$ + то есть на коммунальные услуги приходится $70\\%$ суммы.
+ Шаг 3. Используем второе условие. Подорожание электричества на $50\\%$ — это прибавка $0{,}5\\,Э$, и она равна $10\\%$ от общей суммы: + $$0{,}5\\,Э = 0{,}10\\,S \\implies Э = 0{,}2\\,S,$$ + значит на электричество приходится $20\\%$ суммы.
+ Шаг 4. Доля телефона: + $$Т = S - У - Э = S - 0{,}7\\,S - 0{,}2\\,S = 0{,}1\\,S = 10\\%\\,S.$$ +
Ответ: $10\\%$.
` + }, + { + text: `Определите количество целых решений неравенства + $\\dfrac{2x^2 + 3x - 2}{(2-x)^2(9-x^2)} > 0$.`, + sol: `Числитель: $2x^2 + 3x - 2 = (2x-1)(x+2),$ + корни $x = \\tfrac{1}{2}$ и $x = -2.$
+ Знаменатель: $(2-x)^2(9-x^2) = (2-x)^2(3-x)(3+x).$
+ Множитель $(2-x)^2 \\ge 0$, обращается в $0$ при $x=2$ (исключается); на знак не влияет. + Корни знаменателя: $x = -3,\\; x = 2,\\; x = 3.$
+ Метод интервалов (критические точки $-3,\\,-2,\\,\\tfrac{1}{2},\\,2,\\,3$): + + + + + + + + + +
ИнтервалЗнак
$x<-3$$-$
$(-3;-2)$$+$
$(-2;\\,\\tfrac{1}{2})$$-$
$(\\tfrac{1}{2};\\,2)$$+$
$(2;\\,3)$$-$
$x>3$$-$
+ Решение: $x \\in (-3;\\,-2) \\cup \\left(\\tfrac{1}{2};\\,2\\right).$
+ Целые в этих интервалах: в $(-3;-2)$ — нет; в $(\\tfrac{1}{2};\\,2)$ — это $1.$ + Всего $1$ целое решение. +
Ответ: $1$.
` + }, + { + text: `В прямоугольную трапецию с основаниями $6$ см и $12$ см вписана окружность. + Найдите площадь трапеции.`, + sol: ` + + + + + + + + + + + + + A + B + C + D + 6 + 12 + h=8 + 10 + r=4 + + Свойство описанного четырёхугольника: + суммы противоположных сторон равны.
+ Значит, $6 + 12 = h + b$, где $h$ — перпендикулярная боковая (высота), $b$ — наклонная боковая.
+ Из прямоугольного треугольника с катетами $h$ и $12 - 6 = 6$, гипотенузой $b$:
+ $b = \\sqrt{h^2 + 36}.$
+ Подставляем: $h + \\sqrt{h^2 + 36} = 18 \\implies \\sqrt{h^2 + 36} = 18 - h$
+ $\\implies h^2 + 36 = 324 - 36h + h^2 \\implies 36h = 288 \\implies h = 8$ см.
+ Площадь: + $S = \\dfrac{6 + 12}{2} \\cdot 8 = 9 \\cdot 8 = 72$ см$^2.$ +
Ответ: $72$ см$^2$.
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v53.js b/frontend/js/exam9/variants/v53.js new file mode 100644 index 0000000..b1b91bd --- /dev/null +++ b/frontend/js/exam9/variants/v53.js @@ -0,0 +1,223 @@ +VARIANTS[53] = { + label: "Вариант 53", + tasks: [ + { + text: `Определите, какое из данных чисел НЕ является решением неравенства $x^2 \\leq 4$:`, + opts: [ + ["а", "$0$"], ["б", "$1$"], ["в", "$2$"], ["г", "$-3$"], ["д", "$-2$"], + ], + sol: `Решение неравенства:
+ $x^2 \\leq 4 \\iff |x| \\leq 2 \\iff -2 \\leq x \\leq 2$.
+
+ Проверяем числа: $0,\\ 1,\\ 2,\\ -2$ принадлежат отрезку $[-2;\\ 2]$ — это решения.
+ Число $-3 \\notin [-2;\\ 2]$, так как $(-3)^2 = 9 > 4$.
+
Ответ: г) $-3$
` + }, + { + text: `Значение выражения $\\dfrac{18 \\cdot 9 + 18 \\cdot 3}{18 \\cdot 12}$ равно:`, + opts: [ + ["а", "$18$"], ["б", "$1$"], ["в", "$12$"], ["г", "$2$"], ["д", "$4$"], + ], + sol: `Вычисление: вынесем общий множитель $18$ в числителе:
+ $\\dfrac{18 \\cdot 9 + 18 \\cdot 3}{18 \\cdot 12} = \\dfrac{18(9+3)}{18 \\cdot 12} = \\dfrac{18 \\cdot 12}{18 \\cdot 12} = 1$. +
Ответ: б) $1$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "диагонали прямоугольника равны между собой;"], + ["б", "для прямоугольного треугольника с катетами $m$ и $n$ и гипотенузой $k$ справедливо $m^2 + n^2 = k^2$;"], + ["в", "в треугольнике может быть два прямых угла;"], + ["г", "вписанный угол равен половине соответствующего центрального угла?"], + ], + sol: `Анализ утверждений: +
    +
  • а) верно — свойство прямоугольника;
    • +
  • б) верно — теорема Пифагора;
    • +
  • в) не верно — сумма углов треугольника равна $180^{\\circ}$, а два прямых угла дают $90^{\\circ}+90^{\\circ}=180^{\\circ}$, тогда на третий угол не остаётся;
    • +
  • г) верно — теорема о вписанном угле.
    • +
+
Ответ: в)
` + }, + { + text: `Найдите значения аргумента, при которых значения функции $y = -9x + 7$ неположительны.`, + sol: `Условие: $y \\leq 0$, то есть $-9x + 7 \\leq 0$.
+
+ $-9x \\leq -7 \\iff 9x \\geq 7 \\iff x \\geq \\dfrac{7}{9}$. +
Ответ: $x \\geq \\dfrac{7}{9}$
` + }, + { + text: `Найдите значение выражения $(18n^4 + 27n^3) : (9n^2) - 10n^3 : (5n)$ при $n = -4$.`, + sol: `Правило деления многочлена на одночлен: делим каждый член многочлена на этот одночлен. Для степеней: $\\dfrac{a^m}{a^k}=a^{m-k}$.
+ Шаг 1. Сначала упростим выражение, а потом подставим число — так считать проще, чем сразу подставлять $n=-4$.
+ Шаг 2. Делим первый многочлен на $9n^2$ почленно: + $$\\dfrac{18n^4 + 27n^3}{9n^2} = \\dfrac{18n^4}{9n^2} + \\dfrac{27n^3}{9n^2} = 2n^2 + 3n.$$ + Шаг 3. Делим второй одночлен: + $$\\dfrac{10n^3}{5n} = 2n^2.$$ + Шаг 4. Вычитаем результаты: + $$(2n^2 + 3n) - 2n^2 = 3n.$$ + Шаг 5. Подставляем $n=-4$: + $$3 \\cdot (-4) = -12.$$ +
Ответ: $-12$
` + }, + { + text: `$ABCD$ — равнобедренная трапеция с основаниями $AD = 10$ см, $BC = 6$ см. + Диагональ $BD$ равна $10$ см. Найдите площадь трапеции.`, + sol: ` + + + + + + + A + D + B + C + H + 10 + 6 + h=6 + BD=10 + AH=2 + HD=8 + + Свойство равнобедренной трапеции: высоты, опущенные из вершин меньшего основания на большее, отсекают по краям равные отрезки длины $\\dfrac{AD - BC}{2}$.
+ Теорема Пифагора: $c^2 = a^2 + b^2$ в прямоугольном треугольнике.
+ Формула площади трапеции: $S = \\dfrac{a + b}{2} \\cdot h$.
+ Шаг 1. Находим положение основания высоты.
+ Опустим высоту $BH \\perp AD$. Так как трапеция равнобедренная, + $$AH = \\dfrac{AD - BC}{2} = \\dfrac{10 - 6}{2} = 2\\text{ см},$$ + а $HD = AD - AH = 10 - 2 = 8$ см.
+ Шаг 2. Находим высоту $BH$ по теореме Пифагора.
+ В прямоугольном $\\triangle BHD$ гипотенуза $BD = 10$ и катет $HD = 8$: + $$BH = \\sqrt{BD^2 - HD^2} = \\sqrt{100 - 64} = \\sqrt{36} = 6\\text{ см}.$$ + Шаг 3. Находим площадь. + $$S = \\dfrac{AD + BC}{2} \\cdot BH = \\dfrac{10 + 6}{2} \\cdot 6 = 8 \\cdot 6 = 48\\text{ см}^2.$$ +
Ответ: $48$ см²
` + }, + { + text: `Решите уравнение $1 + \\dfrac{5}{m^2 - m - 6} = \\dfrac{-1}{m + 2}$.`, + sol: `План решения дробного уравнения: разложить знаменатели, найти ОДЗ, умножить на общий знаменатель, решить полученное уравнение и проверить корни.
+ Шаг 1. Раскладываем знаменатель.
+ По теореме Виета подбираем числа $-3$ и $2$ (произведение $-6$, сумма $-1$): + $$m^2 - m - 6 = (m - 3)(m + 2).$$ + Шаг 2. Находим ОДЗ.
+ Знаменатели не должны равняться нулю: $m \\neq 3,\\; m \\neq -2$.
+ Шаг 3. Умножаем обе части на общий знаменатель $(m-3)(m+2)$.
+ $$(m-3)(m+2) + 5 = -(m-3).$$ + Шаг 4. Раскрываем скобки.
+ $m^2 - m - 6 + 5 = -m + 3$;
+ $m^2 - m - 1 = -m + 3$;
+ $m^2 = 4$, откуда $m = \\pm 2$.
+ Шаг 5. Проверяем корни по ОДЗ.
+ $m = -2$ не входит в ОДЗ — отбрасываем. Значит, остаётся $m = 2$. +
Ответ: $m = 2$
` + }, + { + text: `Найдите значение выражения $\\dfrac{9}{2-\\sqrt{13}} - \\dfrac{12}{5+\\sqrt{13}}$. + В ответ запишите число, обратное полученному.`, + sol: `Метод рационализации знаменателя: чтобы убрать корень из знаменателя, умножаем числитель и знаменатель на сопряжённое выражение. При этом используется формула разности квадратов $(a-b)(a+b)=a^2-b^2$.
+ Шаг 1. Преобразуем первую дробь. Сопряжённое к $2-\\sqrt{13}$ — это $2+\\sqrt{13}$: + $$\\dfrac{9}{2-\\sqrt{13}} = \\dfrac{9(2+\\sqrt{13})}{(2-\\sqrt{13})(2+\\sqrt{13})} = \\dfrac{9(2+\\sqrt{13})}{4-13} = \\dfrac{9(2+\\sqrt{13})}{-9} = -(2+\\sqrt{13}).$$ + Шаг 2. Преобразуем вторую дробь. Сопряжённое к $5+\\sqrt{13}$ — это $5-\\sqrt{13}$: + $$\\dfrac{12}{5+\\sqrt{13}} = \\dfrac{12(5-\\sqrt{13})}{25-13} = \\dfrac{12(5-\\sqrt{13})}{12} = 5-\\sqrt{13}.$$ + Шаг 3. Считаем разность дробей: + $$-(2+\\sqrt{13}) - (5-\\sqrt{13}) = -2 - \\sqrt{13} - 5 + \\sqrt{13} = -7.$$ + Шаг 4. По условию записываем число, обратное полученному. Обратное к $-7$ — это + $$\\dfrac{1}{-7} = -\\dfrac{1}{7}.$$ +
Ответ: $-\\dfrac{1}{7}$
` + }, + { + text: `Плиточник планирует уложить $300$ м² плитки. Если он будет укладывать на $5$ м² + в день больше, чем запланировал, то закончит работу на $5$ дней раньше. + Сколько квадратных метров плитки в день планирует укладывать плиточник? + Успеет ли он выполнить заказ за $20$ рабочих дней, если будет работать с опережением? + Ответ обоснуйте.`, + sol: `Пусть $x$ м²/день — плановая производительность ($x>0$).
+ Плановое время: $\\dfrac{300}{x}$ дней; ускоренное: $\\dfrac{300}{x+5}$ дней.
+
+ Уравнение: $\\dfrac{300}{x} - \\dfrac{300}{x+5} = 5$.
+ $300(x+5) - 300x = 5x(x+5)$;
+ $1500 = 5x^2 + 25x \\implies x^2 + 5x - 300 = 0$.
+
+ $D = 25 + 1200 = 1225 = 35^2$;
+ $x = \\dfrac{-5 + 35}{2} = 15$ (отрицательный корень не подходит).
+
+ Плановая производительность: $15$ м²/день; плановый срок $\\dfrac{300}{15}=20$ дней.
+ С опережением: $15+5=20$ м²/день, тогда срок $\\dfrac{300}{20}=15$ дней $< 20$ дней — успеет. +
Ответ: $15$ м²/день; да, успеет (закончит за $15$ дней).
` + }, + { + text: `В треугольник $ABC$ со сторонами $AB = 5$, $BC = 7$, $AC = 8$ вписана окружность. + Касательная $MK$ к окружности пересекает стороны $AB$ и $AC$ в точках $M$ и $K$ + так, что $MK$ не параллельна $BC$. + Найдите периметр треугольника $AMK$.`, + sol: ` + + + + + + + + + + + + + + + + + + + + + + + + + + + A + B + C + + P + Q + T + + M + K + + I + + 5 + 7 + 8 + + AP=AQ=3 + +Шаг 1. Точки касания вписанной окружности. +
Окружность касается стороны $AB$ в точке $P$, стороны $AC$ — в точке $Q$, а касательной $MK$ — в точке $T$. +
Шаг 2. Длина $AP$ через полупериметр. +
Полупериметр: $s=\\dfrac{5+7+8}{2}=10$. +
По известной формуле, касательная из вершины $A$ равна $s$ минус противоположная сторона: +$$AP = AQ = s - BC = 10 - 7 = 3\\text{ см}$$ +
Шаг 3. Касательные из точек $M$ и $K$. +
Из точки $M$ проведены две касательные: одна вдоль $AB$ (касается в $P$), другая — отрезок $MT$. По свойству касательных из одной внешней точки: +$$MP = MT$$ +Аналогично из $K$: +$$KQ = KT$$ +
Шаг 4. Периметр $\\triangle AMK$. +
Распишем периметр и заменим $MT \\to MP$, $TK \\to KQ$: +$$P_{AMK} = \\underbrace{AM}_{\\text{на }AB} + \\underbrace{MK}_{=MT+TK} + \\underbrace{KA}_{\\text{на }AC}$$ +$$= AM + MT + TK + KA = AM + MP + KQ + KA$$ +Группируем по сторонам $AB$ и $AC$: +$$= \\underbrace{(AM + MP)}_{=\\,AP} + \\underbrace{(KQ + KA)}_{=\\,AQ} = AP + AQ$$ +$$= 3 + 3 = 6\\text{ см}$$ +
Ответ: $P_{\\triangle AMK} = 6$ см
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v54.js b/frontend/js/exam9/variants/v54.js new file mode 100644 index 0000000..749c77b --- /dev/null +++ b/frontend/js/exam9/variants/v54.js @@ -0,0 +1,211 @@ +VARIANTS[54] = { + label: "Вариант 54", + tasks: [ + { + text: `Определите, какое из данных чисел НЕ является решением неравенства $x^2 \\leq 9$:`, + opts: [ + ["а", "$0$"], ["б", "$1$"], ["в", "$3$"], ["г", "$-3$"], ["д", "$-8$"], + ], + sol: `Решение неравенства:
+ $x^2 \\leq 9 \\iff |x| \\leq 3 \\iff -3 \\leq x \\leq 3$.
+
+ Проверяем числа: $0,\\ 1,\\ 3,\\ -3$ принадлежат отрезку $[-3;\\ 3]$ — это решения.
+ Число $-8 \\notin [-3;\\ 3]$, так как $(-8)^2 = 64 \\gt 9$.
+
Ответ: д) $-8$
` + }, + { + text: `Значение выражения $\\dfrac{13 \\cdot 2 + 13 \\cdot 6}{13 \\cdot 8}$ равно:`, + opts: [ + ["а", "$13$"], ["б", "$1$"], ["в", "$8$"], ["г", "$2$"], ["д", "$4$"], + ], + sol: `Вычисление: вынесем общий множитель $13$ в числителе:
+ $\\dfrac{13 \\cdot 2 + 13 \\cdot 6}{13 \\cdot 8} = \\dfrac{13(2+6)}{13 \\cdot 8} = \\dfrac{13 \\cdot 8}{13 \\cdot 8} = 1$. +
Ответ: б) $1$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "у ромба все стороны равны между собой;"], + ["б", "если в треугольнике со сторонами $a$, $b$, $c$ выполняется $b^2 + c^2 = a^2$, то треугольник прямоугольный;"], + ["в", "в треугольнике может быть два тупых угла;"], + ["г", "вписанный угол равен половине градусной меры дуги, на которую он опирается?"], + ], + sol: `Анализ утверждений: +
    +
  • а) верно — свойство ромба;
    • +
  • б) верно — обратная теорема Пифагора;
    • +
  • в) не верно — сумма углов треугольника равна $180^{\\circ}$, а два тупых угла в сумме уже превышают $180^{\\circ}$, что невозможно;
    • +
  • г) верно — теорема о вписанном угле.
    • +
+
Ответ: в)
` + }, + { + text: `Найдите значения аргумента, при которых значения функции $y = -7x + 2$ неотрицательны.`, + sol: `Условие: $y \\geq 0$, то есть $-7x + 2 \\geq 0$.
+
+ $-7x \\geq -2 \\iff 7x \\leq 2 \\iff x \\leq \\dfrac{2}{7}$. +
Ответ: $x \\leq \\dfrac{2}{7}$
` + }, + { + text: `Найдите значение выражения $(3m^5 + 4m^3) : m^2 - 15m^4 : (5m)$ при $m = -5$.`, + sol: `Правило деления многочлена на одночлен: делим каждый член многочлена на этот одночлен. Для степеней: $\\dfrac{a^k}{a^l}=a^{k-l}$.
+ Шаг 1. Сначала упростим выражение, а уже потом подставим число — так считать проще.
+ Шаг 2. Делим первый многочлен на $m^2$ почленно: + $$\\dfrac{3m^5 + 4m^3}{m^2} = \\dfrac{3m^5}{m^2}+\\dfrac{4m^3}{m^2} = 3m^3 + 4m.$$ + Шаг 3. Делим второй одночлен: + $$\\dfrac{15m^4}{5m} = 3m^3.$$ + Шаг 4. Вычитаем результаты: + $$(3m^3 + 4m) - 3m^3 = 4m.$$ + Шаг 5. Подставляем $m=-5$: + $$4 \\cdot (-5) = -20.$$ +
Ответ: $-20$
` + }, + { + text: `$ABCD$ — равнобедренная трапеция с основаниями $BC = 2$ см, $AD = 4$ см. + Диагональ $AC$ равна $5$ см. Найдите площадь трапеции.`, + sol: ` + + + + + + + + A + D + B + C + AD=4 + BC=2 + h=4 + AC=5 + + Свойство равнобедренной трапеции: высоты, опущенные из вершин меньшего основания, отсекают на большем основании отрезки длины $\\dfrac{AD - BC}{2}$.
+ Теорема Пифагора: $c^2 = a^2 + b^2$.
+ Формула площади трапеции: $S = \\dfrac{a + b}{2} \\cdot h$.
+ Шаг 1. Опускаем высоту $CH$ из $C$ на $AD$.
+ Так как трапеция равнобедренная, + $$HD = \\dfrac{AD - BC}{2} = \\dfrac{4 - 2}{2} = 1\\text{ см}, \\quad AH = AD - HD = 4 - 1 = 3\\text{ см}.$$ + Шаг 2. Находим высоту $h = CH$ из $\\triangle ACH$.
+ $\\triangle ACH$ прямоугольный (угол $H$ прямой), гипотенуза $AC = 5$, катет $AH = 3$. По теореме Пифагора: + $$h = CH = \\sqrt{AC^2 - AH^2} = \\sqrt{25 - 9} = \\sqrt{16} = 4\\text{ см}.$$ + Шаг 3. Находим площадь. + $$S = \\dfrac{AD + BC}{2} \\cdot h = \\dfrac{4 + 2}{2} \\cdot 4 = 12\\text{ см}^2.$$ +
Ответ: $12$ см²
` + }, + { + text: `Решите уравнение $1 + \\dfrac{7}{n^2 - n - 12} = \\dfrac{-1}{n + 3}$.`, + sol: `План решения дробного уравнения: разложить знаменатели на множители, найти ОДЗ, умножить уравнение на общий знаменатель, решить и проверить корни.
+ Шаг 1. Раскладываем знаменатель.
+ По теореме Виета подбираем числа $-4$ и $3$ (произведение $-12$, сумма $-1$): + $$n^2 - n - 12 = (n - 4)(n + 3).$$ + Шаг 2. ОДЗ: $n \\neq 4,\\; n \\neq -3$.
+ Шаг 3. Умножаем обе части на $(n-4)(n+3)$. + $$(n-4)(n+3) + 7 = -(n-4).$$ + Шаг 4. Раскрываем скобки.
+ $n^2 - n - 12 + 7 = -n + 4$;
+ $n^2 - n - 5 = -n + 4$;
+ $n^2 = 9$, откуда $n = \\pm 3$.
+ Шаг 5. Проверяем по ОДЗ.
+ $n = -3$ не входит в ОДЗ — отбрасываем. Остаётся $n = 3$. +
Ответ: $n = 3$
` + }, + { + text: `Найдите значение выражения $\\dfrac{7}{2-\\sqrt{11}} - \\dfrac{5}{4+\\sqrt{11}}$. + В ответ запишите число, обратное полученному.`, + sol: `Метод рационализации знаменателя: умножаем числитель и знаменатель на сопряжённое выражение, используя формулу разности квадратов $(a-b)(a+b)=a^2-b^2$.
+ Шаг 1. Преобразуем первую дробь. Сопряжённое к $2-\\sqrt{11}$ — это $2+\\sqrt{11}$: + $$\\dfrac{7}{2-\\sqrt{11}} = \\dfrac{7(2+\\sqrt{11})}{(2-\\sqrt{11})(2+\\sqrt{11})} = \\dfrac{7(2+\\sqrt{11})}{4-11} = \\dfrac{7(2+\\sqrt{11})}{-7} = -(2+\\sqrt{11}).$$ + Шаг 2. Преобразуем вторую дробь. Сопряжённое к $4+\\sqrt{11}$ — это $4-\\sqrt{11}$: + $$\\dfrac{5}{4+\\sqrt{11}} = \\dfrac{5(4-\\sqrt{11})}{(4+\\sqrt{11})(4-\\sqrt{11})} = \\dfrac{5(4-\\sqrt{11})}{16-11} = \\dfrac{5(4-\\sqrt{11})}{5} = 4-\\sqrt{11}.$$ + Шаг 3. Находим разность: + $$-(2+\\sqrt{11}) - (4-\\sqrt{11}) = -2 - \\sqrt{11} - 4 + \\sqrt{11} = -6.$$ + Шаг 4. По условию записываем число, обратное полученному: + $$\\dfrac{1}{-6} = -\\dfrac{1}{6}.$$ +
Ответ: $-\\dfrac{1}{6}$
` + }, + { + text: `Плиточник планирует уложить $378$ м² плитки. Если он будет укладывать на $4$ м² + в день больше, чем запланировал, то закончит работу на $6$ дней раньше. + Сколько квадратных метров плитки в день планирует укладывать плиточник? + Успеет ли он выполнить заказ за $21$ рабочий день, если будет работать как запланировал? + Ответ обоснуйте.`, + sol: `Пусть $x$ м²/день — плановая производительность ($x>0$).
+ Плановое время: $\\dfrac{378}{x}$ дней; ускоренное: $\\dfrac{378}{x+4}$ дней.
+
+ Уравнение: $\\dfrac{378}{x} - \\dfrac{378}{x+4} = 6$.
+ $378(x+4) - 378x = 6x(x+4)$;
+ $1512 = 6x^2 + 24x \\implies x^2 + 4x - 252 = 0$.
+
+ $D = 16 + 1008 = 1024 = 32^2$;
+ $x = \\dfrac{-4 + 32}{2} = 14$ (отрицательный корень не подходит).
+
+ Плановая производительность: $14$ м²/день; плановый срок $\\dfrac{378}{14}=27$ дней.
+ Так как $27 > 21$, работая по плану, плиточник не успеет выполнить заказ за $21$ день. +
Ответ: $14$ м²/день; нет, не успеет (закончит за $27$ дней).
` + }, + { + text: `В треугольник $ABC$ со сторонами $AB = 8$, $BC = 10$, $AC = 12$ вписана окружность. + Касательная $MK$ к окружности пересекает стороны $BC$ и $AC$ в точках $M$ и $K$ + так, что $MK$ не параллельна $AB$. + Найдите периметр треугольника $CMK$.`, + sol: ` + + + + + + + + + + + + + + + + + + + + + + + + + A + B + C + + P + Q + T + + M + K + + I + + 8 + 10 + 12 + + CP=CQ=7 + +Шаг 1. Точки касания вписанной окружности.
+ Окружность касается стороны $BC$ в точке $P$, стороны $AC$ — в точке $Q$, а касательной $MK$ — в точке $T$.
+ Шаг 2. Касательная из вершины $C$.
+ Полупериметр: $s=\\dfrac{8+10+12}{2}=15$.
+ По свойству касательных из внешней точки, касательная из $C$ равна $s$ минус противоположная сторона: + $$CP = CQ = s - AB = 15 - 8 = 7\\text{ см.}$$ + Шаг 3. Касательные из $M$ и $K$.
+ Из точки $M$ (на $BC$): $MP = MT$.
+ Из точки $K$ (на $AC$): $KQ = KT$.
+ Шаг 4. Периметр $\\triangle CMK$. + $$P_{CMK} = CM + MK + KC = CM + MT + TK + KC$$ + $$= (CM + MP) + (KQ + KC) = CP + CQ = 7 + 7 = 14\\text{ см.}$$ +
Ответ: $P_{\\triangle CMK} = 14$ см
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v55.js b/frontend/js/exam9/variants/v55.js new file mode 100644 index 0000000..ac62168 --- /dev/null +++ b/frontend/js/exam9/variants/v55.js @@ -0,0 +1,244 @@ +VARIANTS[55] = { + label: "Вариант 55", + tasks: [ + { + text: `Определите, какое из данных выражений НЕ имеет смысла при $x = 0$:`, + opts: [ + ["а", "$\\dfrac{5x}{x^2+49}$"], ["б", "$4\\sqrt{x}-6x$"], ["в", "$\\dfrac{x-6}{2x}$"], + ["г", "$\\dfrac{x}{x-9}$"], ["д", "$\\dfrac{x}{\\sqrt{x}-1}$"], + ], + sol: `Подставим $x=0$ в каждое выражение и проверим, определено ли оно:
    +
  • а) $\\dfrac{5\\cdot 0}{0^2+49}=\\dfrac{0}{49}=0$ — имеет смысл;
    • +
  • б) $4\\sqrt{0}-6\\cdot 0=0$ — имеет смысл;
    • +
  • в) $\\dfrac{0-6}{2\\cdot 0}=\\dfrac{-6}{0}$ — деление на ноль, выражение не определено;
    • +
  • г) $\\dfrac{0}{0-9}=\\dfrac{0}{-9}=0$ — имеет смысл;
    • +
  • д) $\\dfrac{0}{\\sqrt{0}-1}=\\dfrac{0}{-1}=0$ — имеет смысл.
    • +
+
Ответ: в) $\\dfrac{x-6}{2x}$.
` + }, + { + text: `Запись числового выражения $\\dfrac{7^{13}}{7^6 \\cdot 7^5}$ в виде степени с основанием $7$ имеет вид:`, + opts: [ + ["а", "$7^1$"], ["б", "$7^{26}$"], ["в", "$7^3$"], ["г", "$7^4$"], ["д", "$7^5$"], + ], + sol: `Применяем свойства степеней: + $$\\dfrac{7^{13}}{7^6\\cdot 7^5}=\\dfrac{7^{13}}{7^{6+5}}=\\dfrac{7^{13}}{7^{11}}=7^{13-11}=7^2.$$ + Замечание: в предложенных вариантах нет $7^2$. Возможно, в условии опечатка + (например, должно быть $7^{14}$ — тогда $\\dfrac{7^{14}}{7^{11}}=7^3$ и подходит вариант в)). +
Ответ: $7^2$ (в предложенных вариантах отсутствует; вероятно, опечатка — тогда в) $7^3$).
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "прямой угол равен $90^{\\circ}$;"], + ["б", "если из одной точки к прямой проведены перпендикуляр и наклонная, то наклонная меньше перпендикуляра;"], + ["в", "углы при большем основании равнобедренной трапеции равны между собой;"], + ["г", "у квадрата все углы равны между собой?"], + ], + sol: `Проверим каждое утверждение:
    +
  • а) Прямой угол действительно равен $90^{\\circ}$ — верно.
    • +
  • б) Перпендикуляр — это кратчайшее расстояние от точки до прямой, поэтому + перпендикуляр меньше любой наклонной, проведённой из той же точки. + Значит утверждение «наклонная меньше перпендикуляра» — НЕВЕРНО.
    • +
  • в) В равнобедренной трапеции углы при каждом основании равны — верно.
    • +
  • г) У квадрата все углы равны $90^{\\circ}$ — верно.
    • +
+
Ответ: б).
` + }, + { + text: `В библиотеке $12\\%$ всех книг — словари. + Какое количество книг в библиотеке, если словарей $900$?`, + sol: `Пусть всего книг $N$. По условию $12\\%$ от $N$ равно $900$: + $$0{,}12\\cdot N=900\\;\\implies\\; N=\\dfrac{900}{0{,}12}=\\dfrac{900\\cdot 100}{12}=\\dfrac{90000}{12}=7500.$$ +
Ответ: $7500$ книг.
` + }, + { + text: `Постройте график функции $y = \\dfrac{2}{x} + 1$. + Определите целое число, которое не принадлежит области значений функции.`, + sol: `График функции $y=\\dfrac{2}{x}+1$ — гипербола, полученная из $y=\\dfrac{2}{x}$ + сдвигом вверх на $1$. Асимптоты: $x=0$ (вертикальная) и $y=1$ (горизонтальная).
+
+ Найдём область значений. Из равенства $y=1+\\dfrac{2}{x}$ выразим $\\dfrac{2}{x}=y-1$. + Так как $\\dfrac{2}{x}\\neq 0$ (числитель $2\\neq 0$), то $y-1\\neq 0$, то есть $y\\neq 1$.
+ Значит, $E(y)=(-\\infty;1)\\cup(1;+\\infty)$. + Целое число, которое не принадлежит области значений, — это $y=1$. + + + + + + + + + + + + + y = 1 + + + + + + + 1 + + 2 + + −1 + + −2 + + 1 + + 3 + + 2 + + −1 + + x + y + + + (−2; 0) + + (1; 3) + + (2; 2) + + (−1; −1) + +
Ответ: $y=1$.
` + }, + { + text: `Найдите радиус окружности, описанной около правильного треугольника со стороной $a$.`, + sol: `Теорема синусов: в любом треугольнике + $$\\dfrac{a}{\\sin\\alpha} = 2R,$$ + где $a$ — сторона, $\\alpha$ — угол, противолежащий этой стороне, $R$ — радиус описанной окружности.
+ Шаг 1. В правильном (равностороннем) треугольнике все углы равны $60^{\\circ}$, поэтому угол, противолежащий стороне $a$, равен $60^{\\circ}$.
+ Шаг 2. Применяем теорему синусов: + $$\\dfrac{a}{\\sin 60^{\\circ}} = 2R \\implies R = \\dfrac{a}{2 \\sin 60^{\\circ}}.$$ + Шаг 3. Подставляем табличное значение $\\sin 60^{\\circ} = \\dfrac{\\sqrt{3}}{2}$: + $$R = \\dfrac{a}{2 \\cdot \\dfrac{\\sqrt{3}}{2}} = \\dfrac{a}{\\sqrt{3}} = \\dfrac{a\\sqrt{3}}{3}.$$ + (Последний шаг — умножение числителя и знаменателя на $\\sqrt{3}$, чтобы избавиться от иррациональности в знаменателе.) +
+ + + + + + R + A + B + C + O + a + +
+
Ответ: $R=\\dfrac{a\\sqrt{3}}{3}$.
` + }, + { + text: `Решите систему неравенств + $$\\begin{cases} 3 - x > 0, \\\\[4pt] 11x - 3x^2 + 14 \\geq 0 \\end{cases}$$ + и определите количество натуральных решений системы.`, + sol: `Первое неравенство: $3-x>0\\;\\implies\\; x<3$.
+ Второе неравенство: умножим на $-1$ (знак неравенства меняется): + $$11x-3x^2+14\\geq 0\\;\\Leftrightarrow\\; 3x^2-11x-14\\leq 0.$$ + Корни уравнения $3x^2-11x-14=0$: $D=121+168=289=17^2$, + $$x=\\dfrac{11\\pm 17}{6}\\;\\implies\\; x_1=-1,\\quad x_2=\\dfrac{28}{6}=\\dfrac{14}{3}.$$ + Парабола ветвями вверх, поэтому $3x^2-11x-14\\leq 0$ при $-1\\leq x\\leq \\dfrac{14}{3}$.
+
+ Пересечение: $-1\\leq x<3$.
+ Натуральные решения (целые $\\geq 1$): $x=1,\\;2$. Их 2. +
Ответ: $x\\in[-1;3)$; натуральных решений $2$ (числа $1$ и $2$).
` + }, + { + text: `Строительный подрядчик планирует купить $4$ т $300$ кг одинарного облицовочного кирпича. + Масса одного кирпича — $4{,}3$ кг. Во сколько рублей обойдётся наиболее дешёвый вариант + покупки у одного из трёх поставщиков:
+
+ + + + + +
ПоставщикЦена, р./шт.Доставка, р.Условие
А1,70700—
Б1,80600При заказе свыше 1500 р. — доставка бесплатно
В1,90500При заказе свыше 1600 р. — доставка со скидкой 50%
`, + sol: `Количество кирпичей: $4$ т $300$ кг $=4300$ кг. + $$N=\\dfrac{4300}{4{,}3}=1000\\text{ штук}.$$ + Поставщик А: $1000\\cdot 1{,}70+700=1700+700=\\mathbf{2400}$ р.
+ Поставщик Б: стоимость кирпича $1000\\cdot 1{,}80=1800$ р. Так как $1800>1500$, + доставка бесплатна. Итого $\\mathbf{1800}$ р.
+ Поставщик В: $1000\\cdot 1{,}90=1900$ р. Так как $1900>1600$, доставка со скидкой $50\\%$: + $\\dfrac{500}{2}=250$ р. Итого $1900+250=\\mathbf{2150}$ р.
+
+ Сравниваем: $1800<2150<2400$. Самый дешёвый — поставщик Б. +
Ответ: $1800$ р. (поставщик Б).
` + }, + { + text: `Решите уравнение + $\\dfrac{6}{x^2-36} + \\dfrac{1}{x^2-12x+36} + \\dfrac{1}{2x+12} = 0$.`, + sol: `Формулы: разность квадратов $a^2 - b^2 = (a-b)(a+b)$; квадрат разности $a^2 - 2ab + b^2 = (a-b)^2$.
+ Шаг 1. Раскладываем знаменатели. + $$x^2 - 36 = (x-6)(x+6),\\quad x^2 - 12x + 36 = (x-6)^2,\\quad 2x + 12 = 2(x+6).$$ + Шаг 2. Находим ОДЗ.
+ Знаменатели не должны быть равны нулю: $x \\neq 6,\\; x \\neq -6$.
+ Шаг 3. Умножаем обе части уравнения на общий знаменатель $2(x-6)^2(x+6)$.
+ Получаем (после сокращений): + $$12(x-6) + 2(x+6) + (x-6)^2 = 0.$$ + Шаг 4. Раскрываем скобки. + $$12x - 72 + 2x + 12 + x^2 - 12x + 36 = 0,$$ + $$x^2 + 2x - 24 = 0.$$ + Шаг 5. Решаем квадратное уравнение.
+ Дискриминант: $D = 4 + 96 = 100$, $\\sqrt{D} = 10$. + $$x = \\dfrac{-2 \\pm 10}{2} \\implies x_1 = 4,\\; x_2 = -6.$$ + Шаг 6. Проверяем по ОДЗ.
+ $x = -6$ не входит в ОДЗ — отбрасываем. Остаётся $x = 4$. +
Ответ: $x=4$.
` + }, + { + text: `Основания трапеции равны $5$ см и $10$ см, диагонали трапеции равны $13$ см и $14$ см. + Найдите площадь трапеции.`, + sol: `Пусть в трапеции $ABCD$: $BC=5$, $AD=10$, $AC=13$, $BD=14$.
+ Метод параллельного переноса диагонали. Перенесём диагональ $BD$ параллельно на вектор + $\\overrightarrow{BC}$: точка $B$ перейдёт в $C$, точка $D$ — в новую точку $D'$ на прямой $AD$. + Тогда $CD'=BD=14$ и $DD'=BC=5$, поэтому + $$AD'=AD+DD'=10+5=15.$$ + В треугольнике $ACD'$ известны все три стороны: $AC=13$, $CD'=14$, $AD'=15$.
+ Площадь трапеции равна площади треугольника $ACD'$ (так как $BCDD'$ — параллелограмм, + и площадь $\\triangle BCD$ равна площади $\\triangle DCD'$).
+ По формуле Герона: $p=\\dfrac{13+14+15}{2}=21$, + $$S=\\sqrt{21\\cdot(21-13)\\cdot(21-14)\\cdot(21-15)}=\\sqrt{21\\cdot 8\\cdot 7\\cdot 6}=\\sqrt{7056}=84.$$ + + + + + + + + + + + + + + + + A + B + C + D + D' + + AD=10 + BC=5 + AC=13 + BD=14 + CD'=14 + DD'=5 + △ACD' (13-14-15) + +
Ответ: $S=84$ см².
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v56.js b/frontend/js/exam9/variants/v56.js new file mode 100644 index 0000000..5dc9b39 --- /dev/null +++ b/frontend/js/exam9/variants/v56.js @@ -0,0 +1,211 @@ +VARIANTS[56] = { + label: "Вариант 56", + tasks: [ + { + text: `Определите, какое из данных выражений НЕ имеет смысла при $n = 0$:`, + opts: [ + ["а", "$\\dfrac{n}{n+7}$"], ["б", "$19n-4\\sqrt{n}$"], ["в", "$\\dfrac{n-6}{n^2+36}$"], + ["г", "$\\dfrac{n+1}{3n}$"], ["д", "$\\dfrac{n}{\\sqrt{n}-1}$"], + ], + sol: `Подставим $n=0$ в каждое выражение и проверим, определено ли оно:
    +
  • а) $\\dfrac{0}{0+7}=\\dfrac{0}{7}=0$ — имеет смысл;
    • +
  • б) $19\\cdot 0-4\\sqrt{0}=0$ — имеет смысл;
    • +
  • в) $\\dfrac{0-6}{0^2+36}=\\dfrac{-6}{36}=-\\dfrac{1}{6}$ — имеет смысл;
    • +
  • г) $\\dfrac{0+1}{3\\cdot 0}=\\dfrac{1}{0}$ — деление на ноль, выражение не определено;
    • +
  • д) $\\dfrac{0}{\\sqrt{0}-1}=\\dfrac{0}{-1}=0$ — имеет смысл.
    • +
+
Ответ: г) $\\dfrac{n+1}{3n}$.
` + }, + { + text: `Запись числового выражения $\\dfrac{5^{16} \\cdot 5^4}{5^{18}}$ в виде степени с основанием $5$ имеет вид:`, + opts: [ + ["а", "$5^1$"], ["б", "$5^2$"], ["в", "$5^{38}$"], ["г", "$5^4$"], ["д", "$5^3$"], + ], + sol: `Применяем свойства степеней: + $$\\dfrac{5^{16}\\cdot 5^4}{5^{18}}=\\dfrac{5^{16+4}}{5^{18}}=\\dfrac{5^{20}}{5^{18}}=5^{20-18}=5^2.$$ +
Ответ: б) $5^2$.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "развёрнутый угол равен $180^{\\circ}$;"], + ["б", "если из одной точки к прямой проведены перпендикуляр и наклонная, то перпендикуляр меньше наклонной;"], + ["в", "у любого ромба диагонали равны;"], + ["г", "диагонали равнобедренной трапеции равны между собой?"], + ], + sol: `Проверим каждое утверждение:
    +
  • а) Развёрнутый угол равен $180^{\\circ}$ — верно.
    • +
  • б) Перпендикуляр — кратчайшее расстояние от точки до прямой, поэтому он меньше любой наклонной — верно.
    • +
  • в) У ромба диагонали вообще говоря не равны между собой (они равны лишь у квадрата — частного случая ромба). Это утверждение НЕВЕРНО.
    • +
  • г) У равнобедренной трапеции диагонали равны — верно.
    • +
+
Ответ: в).
` + }, + { + text: `На республиканском субботнике учащиеся высадили кустарники, среди которых — $10$ кустов сирени. + Какое количество кустарников было высажено, если число кустов сирени составило $8\\%$ + всех высаженных кустарников?`, + sol: `Пусть всего кустарников $N$. По условию $8\\%$ от $N$ равно $10$: + $$0{,}08\\cdot N=10\\;\\implies\\; N=\\dfrac{10}{0{,}08}=\\dfrac{10\\cdot 100}{8}=\\dfrac{1000}{8}=125.$$ +
Ответ: $125$ кустарников.
` + }, + { + text: `Постройте график функции $y = \\dfrac{4}{x} - 1$. + Определите целое число, которое не принадлежит области значений функции.`, + sol: `График функции $y=\\dfrac{4}{x}-1$ — гипербола, полученная из $y=\\dfrac{4}{x}$ + сдвигом вниз на $1$. Асимптоты: $x=0$ (вертикальная) и $y=-1$ (горизонтальная).
+
+ Из равенства $y=-1+\\dfrac{4}{x}$ выразим $\\dfrac{4}{x}=y+1$. + Так как $\\dfrac{4}{x}\\neq 0$, то $y+1\\neq 0$, то есть $y\\neq -1$.
+ Значит, $E(y)=(-\\infty;-1)\\cup(-1;+\\infty)$. + Целое число, которое не принадлежит области значений, — это $y=-1$. + + + + + + + + + + y = -1 + + + + 1 + + -1 + + 1 + + 2 + x + y + + (1; 3) + + (2; 1) + +
Ответ: $y=-1$.
` + }, + { + text: `В правильный треугольник со стороной $a$ вписана окружность. Найдите радиус окружности.`, + sol: `Формула радиуса вписанной окружности: + $$r = \\dfrac{S}{s},$$ + где $S$ — площадь треугольника, $s$ — полупериметр.
+ Шаг 1. В равностороннем треугольнике со стороной $a$ все углы равны $60^{\\circ}$.
+ Шаг 2. Находим площадь.
+ По формуле площади равностороннего треугольника: + $$S = \\dfrac{\\sqrt{3}}{4}a^2.$$ + Шаг 3. Находим полупериметр.
+ Все три стороны равны $a$, поэтому + $$s = \\dfrac{3a}{2}.$$ + Шаг 4. Подставляем в формулу радиуса. + $$r = \\dfrac{S}{s} = \\dfrac{\\dfrac{\\sqrt{3}}{4}a^2}{\\dfrac{3a}{2}} = \\dfrac{\\sqrt{3}\\,a^2}{4} \\cdot \\dfrac{2}{3a} = \\dfrac{a\\sqrt{3}}{6}.$$ +
+ + + + + + r + A + B + C + O + a + +
+
Ответ: $r=\\dfrac{a\\sqrt{3}}{6}$.
` + }, + { + text: `Решите систему неравенств + $$\\begin{cases} 2 - x > 0, \\\\[4pt] 5x - 2x^2 + 7 \\geq 0 \\end{cases}$$ + и определите количество натуральных решений системы.`, + sol: `Первое неравенство: $2-x>0\\;\\implies\\; x<2$.
+ Второе неравенство: $5x-2x^2+7\\geq 0\\;\\Leftrightarrow\\; 2x^2-5x-7\\leq 0$.
+ Корни уравнения $2x^2-5x-7=0$: $D=25+56=81=9^2$, + $$x=\\dfrac{5\\pm 9}{4}\\;\\implies\\; x_1=\\dfrac{5-9}{4}=-1,\\quad x_2=\\dfrac{5+9}{4}=\\dfrac{14}{4}=\\dfrac{7}{2}.$$ + Парабола ветвями вверх, поэтому $2x^2-5x-7\\leq 0$ при $-1\\leq x\\leq \\dfrac{7}{2}$.
+
+ Пересечение: $\\left[-1;\\dfrac{7}{2}\\right)\\cap(-\\infty;2)=\\left[-1;2\\right)$.
+ Натуральные решения (целые $\\geq 1$): $x=1$. Их 1. +
Ответ: $x\\in[-1;2)$; натуральных решений $1$ (число $1$).
` + }, + { + text: `Строительный подрядчик планирует купить $3$ т $600$ кг клинкерного кирпича. + Масса одного кирпича — $3{,}6$ кг. Во сколько рублей обойдётся наиболее дешёвый вариант + покупки у одного из трёх поставщиков:
+
+ + + + + +
ПоставщикЦена, р./шт.Доставка, р.Условие
А2,7700—
Б2,8600При заказе свыше 3500 р. — доставка бесплатно
В2,9500При заказе свыше 3600 р. — доставка со скидкой 50%
`, + sol: `Количество кирпичей: $3$ т $600$ кг $=3600$ кг. + $$N=\\dfrac{3600}{3{,}6}=1000\\text{ штук}.$$ + Поставщик А: $1000\\cdot 2{,}7+700=2700+700=\\mathbf{3400}$ р.
+ Поставщик Б: стоимость кирпича $1000\\cdot 2{,}8=2800$ р. Так как $2800<3500$, + условие бесплатной доставки не выполнено. Итого $2800+600=\\mathbf{3400}$ р.
+ Поставщик В: $1000\\cdot 2{,}9=2900$ р. Так как $2900<3600$, + скидка на доставку не действует. Итого $2900+500=\\mathbf{3400}$ р.
+
+ Все три поставщика дают одинаковую стоимость $3400$ р. Наиболее дешёвый вариант — $3400$ р. (любой из поставщиков). +
Ответ: $3400$ р. (все три поставщика равноценны).
` + }, + { + text: `Решите уравнение + $\\dfrac{1}{x^2-9} + \\dfrac{1}{x^2-6x+9} = \\dfrac{1}{2x^2+6x}$.`, + sol: `Формулы: разность квадратов $a^2 - b^2 = (a-b)(a+b)$; квадрат разности $a^2 - 2ab + b^2 = (a-b)^2$.
+ Шаг 1. Раскладываем знаменатели. + $$x^2 - 9 = (x-3)(x+3),\\quad x^2 - 6x + 9 = (x-3)^2,\\quad 2x^2 + 6x = 2x(x+3).$$ + Шаг 2. Находим ОДЗ.
+ Знаменатели не равны нулю: $x \\neq 3,\\; x \\neq -3,\\; x \\neq 0$.
+ Шаг 3. Умножаем обе части уравнения на общий знаменатель $2x(x-3)^2(x+3)$.
+ Получаем: + $$2x(x-3) + 2x(x+3) = (x-3)(x+3).$$ + Шаг 4. Раскрываем скобки. + $$2x^2 - 6x + 2x^2 + 6x = x^2 - 9,$$ + $$4x^2 = x^2 - 9,$$ + $$3x^2 = -9, \\quad x^2 = -3.$$ + Шаг 5. Анализируем результат.
+ Квадрат любого действительного числа $\\geq 0$, поэтому равенство $x^2 = -3$ невозможно. +
Ответ: уравнение не имеет решений.
` + }, + { + text: `Основания трапеции равны $7$ см и $14$ см, диагонали трапеции равны $13$ см и $20$ см. + Найдите площадь трапеции.`, + sol: `Пусть в трапеции $ABCD$: $BC=7$, $AD=14$, $AC=13$, $BD=20$.
+ Метод параллельного переноса диагонали. Перенесём диагональ $BD$ параллельно на вектор + $\\overrightarrow{BC}$: точка $B$ перейдёт в $C$, точка $D$ — в новую точку $D'$ на прямой $AD$. + Тогда $CD'=BD=20$ и $DD'=BC=7$, поэтому + $$AD'=AD+DD'=14+7=21.$$ + В треугольнике $ACD'$ известны все три стороны: $AC=13$, $CD'=20$, $AD'=21$.
+ Площадь трапеции равна площади треугольника $ACD'$.
+ По формуле Герона: $p=\\dfrac{13+20+21}{2}=27$, + $$S=\\sqrt{27\\cdot(27-13)\\cdot(27-20)\\cdot(27-21)}=\\sqrt{27\\cdot 14\\cdot 7\\cdot 6}=\\sqrt{15876}=126.$$ + + + + + + + + A + B + C + D + D' + AD=14 + BC=7 + AC=13 + BD=20 + CD'=20 + DD'=7 + △ACD'(13-20-21) + +
Ответ: $S=126$ см².
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v57.js b/frontend/js/exam9/variants/v57.js new file mode 100644 index 0000000..2058cc4 --- /dev/null +++ b/frontend/js/exam9/variants/v57.js @@ -0,0 +1,219 @@ +VARIANTS[57] = { + label: "Вариант 57", + tasks: [ + { + text: `Определите рисунок, на котором изображён график функции $y = 3$:`, + figure: ``, + sol: `Уравнение $y = 3$ задаёт постоянную функцию: при любом значении $x$ значение $y$ равно $3$. +
Графиком является горизонтальная прямая, параллельная оси $Ox$ и проходящая через точку $(0;\\,3)$ (на высоте $3$ над осью абсцисс). + + + + + x + y + 0 + + + y = 3 + + 3 + +
Ответ: горизонтальная прямая $y=3$, параллельная оси $Ox$ и проходящая через точку $(0;\\,3)$.
` + }, + { + text: `Определите, какой из данных одночленов записан в стандартном виде:`, + opts: [ + ["а", "$3xyuz$"], ["б", "$-x \\cdot \\dfrac{1}{2} \\cdot y \\cdot z$"], ["в", "$0{,}25x^5yz$"], + ["г", "$0{,}5x^5y \\cdot 2z$"], ["д", "$x^5y \\cdot 2zy$"], + ], + sol: `Одночлен записан в стандартном виде, если он представлен как произведение одного числового коэффициента и переменных, каждая из которых встречается ровно один раз и возведена в натуральную степень. +
    +
  • а) $3xyuz$ — коэффициент один ($3$), но обычно требуется упорядочение; формально допустимо, однако чаще считают нестандартным из-за порядка переменных;
    • +
  • б) $-x\\cdot\\dfrac{1}{2}\\cdot y\\cdot z$ — два числовых множителя ($-1$ и $\\tfrac{1}{2}$);
    • +
  • в) $0{,}25x^5yz$ — один коэффициент $0{,}25$, переменные $x^5,\\ y,\\ z$ записаны по одному разу — стандартный вид ✓
    • +
  • г) $0{,}5x^5y\\cdot 2z$ — два числовых множителя ($0{,}5$ и $2$);
    • +
  • д) $x^5y\\cdot 2zy$ — переменная $y$ встречается дважды.
    • +
+
Ответ: в) $0{,}25x^5yz$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "$\\sin 30^{\\circ} = \\dfrac{1}{2}$;"], + ["б", "площадь треугольника равна половине произведения двух его сторон на синус угла между ними;"], + ["в", "сумма углов прямоугольника равна $270^{\\circ}$;"], + ["г", "периметр квадрата со стороной $a$ равен $4a$?"], + ], + sol: `
    +
  • а) $\\sin 30^{\\circ}=\\dfrac{1}{2}$ — табличное значение, верно;
    • +
  • б) $S_{\\triangle}=\\dfrac{1}{2}ab\\sin\\gamma$ — верно;
    • +
  • в) Сумма углов любого выпуклого четырёхугольника, в том числе прямоугольника, равна $360^{\\circ}$ ($4\\cdot 90^{\\circ}=360^{\\circ}$), а не $270^{\\circ}$ — НЕВЕРНО;
    • +
  • г) $P_{\\text{кв}}=4a$ — верно.
    • +
+
Ответ: в)
` + }, + { + text: `Известно, что функция $y = f(x)$ нечётная и $f(5) = -9$, $f(-3) = 4$. + Найдите значение выражения $f(3) + f(-5)$.`, + sol: `Для нечётной функции выполняется тождество $f(-x)=-f(x)$. +
Тогда: +$$f(-5) = -f(5) = -(-9) = 9,$$ +$$f(3) = -f(-3) = -4.$$ +Складываем: +$$f(3)+f(-5) = -4 + 9 = 5.$$ +
Ответ: $5$
` + }, + { + text: `Коробка конфет кондитерской фабрики «Коммунарка» стоит $17$ р. $60$ к. + Какое наибольшее количество коробок можно купить на $130$ р.?`, + sol: `Метод составления неравенства по условию: общая стоимость покупки не должна превосходить имеющейся суммы.
+Шаг 1. Переведём цену в рубли: $17$ р. $60$ к. $= 17{,}60$ р.
+Шаг 2. Обозначим $n$ — число коробок. Тогда стоимость $n$ коробок равна $17{,}60\\,n$ рублей. По условию её хватает на $130$ р., поэтому +$$17{,}60\\,n \\leq 130.$$ +Шаг 3. Решаем неравенство, делим обе части на положительное число $17{,}60$: +$$n \\leq \\dfrac{130}{17{,}60} = 7{,}3863\\ldots$$ +Шаг 4. Так как $n$ — натуральное число (количество коробок), наибольшее значение, удовлетворяющее $n\\leq 7{,}38\\ldots$, — это $n=7$.
+Шаг 5. Проверим: +
    +
  • $7\\cdot 17{,}60 = 123{,}20$ р. $\\leq 130$ р. — подходит;
    • +
  • $8\\cdot 17{,}60 = 140{,}80$ р. $\\gt 130$ р. — не подходит.
    • +
+
Ответ: $7$ коробок
` + }, + { + text: `В квадрат, диагональ которого равна $8$ см, вписана окружность. + Найдите длину этой окружности.`, + sol: `Теорема Пифагора: $c^2 = a^2 + b^2$.
+Свойство квадрата: диагональ квадрата делит его на два равных прямоугольных равнобедренных треугольника, у которых катеты — стороны квадрата.
+Свойство вписанной окружности: радиус окружности, вписанной в квадрат, равен половине стороны.
+Формула длины окружности: $C = 2\\pi R$.
+Шаг 1. Находим сторону квадрата.
+Пусть $a$ — сторона, $d = 8$ — диагональ. По теореме Пифагора в одном из треугольников (катеты $a$ и $a$): +$$d^2 = a^2 + a^2 = 2a^2 \\implies 2a^2 = 64 \\implies a^2 = 32 \\implies a = \\sqrt{32} = 4\\sqrt{2}\\text{ см}.$$ +Шаг 2. Находим радиус вписанной окружности. +$$R = \\dfrac{a}{2} = \\dfrac{4\\sqrt{2}}{2} = 2\\sqrt{2}\\text{ см}.$$ +Шаг 3. Находим длину окружности. +$$C = 2\\pi R = 2\\pi \\cdot 2\\sqrt{2} = 4\\pi\\sqrt{2}\\text{ см}.$$ + + + + + + + A + B + C + D + R + O + a = 4√2 + +
Ответ: $C = 4\\pi\\sqrt{2}$ см
` + }, + { + text: `Найдите наименьшее целое значение переменной, при котором сумма дробей + $\\dfrac{2x-5}{4}$ и $\\dfrac{3-4x}{6}$ неположительна.`, + sol: `Метод решения линейного неравенства с дробями: приводим к общему знаменателю, домножаем на положительное число (знак не меняется), решаем.
+Шаг 1. «Неположительна» означает «не больше нуля», то есть $\\leq 0$. Записываем неравенство: +$$\\dfrac{2x-5}{4} + \\dfrac{3-4x}{6} \\leq 0.$$ +Шаг 2. Приводим к общему знаменателю $12$ (наименьшее общее кратное чисел $4$ и $6$): +$$\\dfrac{3(2x-5) + 2(3-4x)}{12} \\leq 0.$$ +Шаг 3. Раскрываем скобки в числителе: +$$\\dfrac{6x-15+6-8x}{12} \\leq 0 \\iff \\dfrac{-2x-9}{12} \\leq 0.$$ +Шаг 4. Так как знаменатель $12\\gt 0$, неравенство равносильно тому, что числитель $\\leq 0$: +$$-2x-9 \\leq 0 \\iff -2x \\leq 9 \\iff x \\geq -\\dfrac{9}{2} = -4{,}5$$ +(при делении на отрицательное число $-2$ знак неравенства меняется на противоположный).
+Шаг 5. Ищем наименьшее целое $x$, удовлетворяющее $x\\geq -4{,}5$. Это $x=-4$. +
Ответ: $-4$
` + }, + { + text: `Упростите выражение + $\\left(\\dfrac{9}{4-\\sqrt{7}} - \\dfrac{33}{6-\\sqrt{3}} - \\dfrac{4}{\\sqrt{7}+\\sqrt{3}}\\right)^{\\!2}$.`, + sol: `Метод рационализации знаменателя: чтобы убрать корень из знаменателя, умножаем числитель и знаменатель на сопряжённое выражение. При этом срабатывает формула разности квадратов $(a-b)(a+b)=a^2-b^2$.
+Шаг 1. Преобразуем первую дробь. Сопряжённое к $4-\\sqrt{7}$ — это $4+\\sqrt{7}$: +$$\\dfrac{9}{4-\\sqrt{7}} = \\dfrac{9(4+\\sqrt{7})}{(4-\\sqrt{7})(4+\\sqrt{7})} = \\dfrac{9(4+\\sqrt{7})}{16-7} = \\dfrac{9(4+\\sqrt{7})}{9} = 4+\\sqrt{7}.$$ +Шаг 2. Преобразуем вторую дробь. Сопряжённое к $6-\\sqrt{3}$ — это $6+\\sqrt{3}$: +$$\\dfrac{33}{6-\\sqrt{3}} = \\dfrac{33(6+\\sqrt{3})}{36-3} = \\dfrac{33(6+\\sqrt{3})}{33} = 6+\\sqrt{3}.$$ +Шаг 3. Преобразуем третью дробь. Сопряжённое к $\\sqrt{7}+\\sqrt{3}$ — это $\\sqrt{7}-\\sqrt{3}$: +$$\\dfrac{4}{\\sqrt{7}+\\sqrt{3}} = \\dfrac{4(\\sqrt{7}-\\sqrt{3})}{7-3} = \\dfrac{4(\\sqrt{7}-\\sqrt{3})}{4} = \\sqrt{7}-\\sqrt{3}.$$ +Шаг 4. Подставляем в исходное выражение и приводим подобные слагаемые: +$$(4+\\sqrt{7}) - (6+\\sqrt{3}) - (\\sqrt{7}-\\sqrt{3}) = 4+\\sqrt{7} - 6 - \\sqrt{3} - \\sqrt{7} + \\sqrt{3} = -2.$$ +Шаг 5. По условию надо возвести в квадрат: +$$(-2)^2 = 4.$$ +
Ответ: $4$
` + }, + { + text: `Дан параллелограмм $ABCD$. На стороне $BC$ взята некоторая точка $M$. + Отрезок $DM$ пересекает диагональ $AC$ в точке $K$. + Площади треугольников $MCK$ и $DCK$ равны соответственно $9$ см² и $15$ см². + Найдите площадь параллелограмма.`, + sol: ` + + + + + + + + + + + + + + + A + B + C + D + M + K + 9 + 15 + +Треугольники $MCK$ и $DCK$ имеют общую вершину $C$, а основания $MK$ и $KD$ лежат на одной прямой $DM$. Значит, их площади относятся как длины оснований: +$$\\dfrac{S_{MCK}}{S_{DCK}} = \\dfrac{MK}{KD} = \\dfrac{9}{15} = \\dfrac{3}{5}.$$ +Так как $AD\\parallel BC$, то $\\triangle AKD \\sim \\triangle CKM$ (по двум углам: $\\angle AKD=\\angle CKM$ — вертикальные, $\\angle KAD=\\angle KCM$ — накрест-лежащие). +
Коэффициент подобия: +$$\\dfrac{AD}{CM} = \\dfrac{KD}{KM} = \\dfrac{5}{3} \\implies CM = \\dfrac{3}{5}AD.$$ +Площадь треугольника $CDM$: +$$S_{CDM} = S_{MCK} + S_{DCK} = 9 + 15 = 24\\text{ см}^2.$$ +С другой стороны, если $h$ — высота параллелограмма, опущенная на $BC$, то $S_{CDM}=\\dfrac{1}{2}\\cdot CM\\cdot h$: +$$24 = \\dfrac{1}{2}\\cdot\\dfrac{3}{5}AD\\cdot h \\implies AD\\cdot h = \\dfrac{24\\cdot 2\\cdot 5}{3} = 80.$$ +Площадь параллелограмма равна произведению стороны на высоту: +$$S_{ABCD} = AD\\cdot h = 80\\text{ см}^2.$$ +
Ответ: $80$ см²
` + }, + { + text: `Решите систему уравнений + $$\\begin{cases} xy - x - y = 29, \\\\[4pt] x^2 + y^2 - x - y = 72. \\end{cases}$$`, + sol: `Идея: в каждом уравнении встречается выражение $x+y$. Введём одну замену $s = x+y$. +

+Шаг 1. Из первого уравнения выражаем $xy$: +$$xy = s + 29$$ +Из второго — выражаем $x^2+y^2$: +$$x^2+y^2 = s + 72$$ +Шаг 2. Применяем тождество $(x+y)^2 = x^2+2xy+y^2$, то есть $x^2+y^2 = s^2 - 2xy$. +
Подставляем найденные выражения: +$$s + 72 = s^2 - 2(s+29)$$ +$$s + 72 = s^2 - 2s - 58$$ +$$s^2 - 3s - 130 = 0$$ +Шаг 3. Решаем квадратное уравнение. +
$D = 9+520 = 529 = 23^2$, корни: +$$s = \\dfrac{3\\pm23}{2}: \\quad s_1 = 13,\\quad s_2 = -10$$ +Шаг 4. Для каждого $s$ восстанавливаем $x$ и $y$. +
Если $x+y = s$ и $xy = s+29$, то по обратной теореме Виета $x$ и $y$ — корни уравнения $t^2 - st + (s+29) = 0$. +

+Случай 1: $s = 13$, $xy = 42$. +$$t^2 - 13t + 42 = 0 \\implies (t-6)(t-7)=0 \\implies t = 6\\text{ или }7$$ +$$(x;\\,y) = (6;\\,7)\\ \\text{или}\\ (7;\\,6)$$ +Проверка: $6\\cdot7-6-7 = 42-13 = 29$ ✓;  $36+49-6-7 = 72$ ✓. +

+Случай 2: $s = -10$, $xy = 19$. +$$t^2 + 10t + 19 = 0 \\implies D = 100-76 = 24 \\implies t = -5\\pm\\sqrt{6}$$ +$$(x;\\,y) = (-5+\\sqrt{6};\\,-5-\\sqrt{6})\\ \\text{или}\\ (-5-\\sqrt{6};\\,-5+\\sqrt{6})$$ +
Ответ: $(6;\\,7),\\ (7;\\,6),\\ (-5+\\sqrt{6};\\,-5-\\sqrt{6}),\\ (-5-\\sqrt{6};\\,-5+\\sqrt{6})$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v58.js b/frontend/js/exam9/variants/v58.js new file mode 100644 index 0000000..5083e4b --- /dev/null +++ b/frontend/js/exam9/variants/v58.js @@ -0,0 +1,172 @@ +VARIANTS[58] = { + label: "Вариант 58", + tasks: [ + { + text: `Определите рисунок, на котором изображён график функции $y = -2$:`, + figure: ``, + sol: `Уравнение $y = -2$ задаёт постоянную функцию. Граф — горизонтальная прямая через $(0;\\,-2)$.
Ответ: горизонтальная прямая $y=-2$, параллельная оси $Ox$, проходящая через $(0;\\,-2)$.
` + }, + { + text: `Определите, какой из данных одночленов записан в стандартном виде:`, + opts: [ + ["а", "$-a \\cdot \\dfrac{1}{4} \\cdot b \\cdot c$"], ["б", "$5abcc$"], ["в", "$0{,}5a^7bc^2$"], + ["г", "$0{,}35a^3b \\cdot 2c$"], ["д", "$a^4b \\cdot 2cb$"], + ], + sol: `Стандартный вид: один коэффициент, каждая переменная один раз. в) $0{,}5a^7bc^2$ — ✓
Ответ: в) $0{,}5a^7bc^2$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "$\\cos 30^{\\circ} = \\dfrac{\\sqrt{3}}{2}$;"], + ["б", "площадь параллелограмма равна произведению соседних сторон на синус угла между ними;"], + ["в", "сумма углов прямоугольника равна $300^{\\circ}$;"], + ["г", "площадь квадрата со стороной $a$ равна $a^2$?"], + ], + sol: `Сумма углов четырёхугольника $=360^{\\circ}$, а не $300^{\\circ}$ — НЕВЕРНО.
Ответ: в)
` + }, + { + text: `Известно, что функция $y = f(x)$ нечётная и $f(-3) = 4$, $f(-5) = -8$. + Найдите значение выражения $f(3) + f(5)$.`, + sol: `Для нечётной функции $f(-x)=-f(x)$: $f(3)=-f(-3)=-4$; $f(5)=-f(-5)=8$. Сумма: $-4+8=4$.
Ответ: $4$
` + }, + { + text: `Коробка конфет кондитерской фабрики «Спартак» стоит $15$ р. $50$ к. + Какое наибольшее количество коробок можно купить на $150$ р.?`, + sol: `Метод составления неравенства по условию: общая стоимость покупки не должна превосходить имеющейся суммы.
+Шаг 1. Переводим цену в рубли: $15$ р. $50$ к. $= 15{,}50$ р.
+Шаг 2. Пусть $n$ — число коробок. Стоимость $n$ коробок равна $15{,}50\\,n$ рублей. По условию её хватает на $150$ р., значит +$$15{,}50\\,n \\leq 150.$$ +Шаг 3. Делим обе части на положительное число $15{,}50$: +$$n \\leq \\dfrac{150}{15{,}50} = 9{,}677\\ldots$$ +Шаг 4. Так как $n$ — натуральное (количество коробок), наибольшее значение $n=9$.
+Шаг 5. Проверим: $9\\cdot 15{,}50 = 139{,}50$ р. $\\leq 150$ р. — подходит; а $10\\cdot 15{,}50 = 155$ р. $\\gt 150$ р. — не подходит. +
Ответ: $9$ коробок
` + }, + { + text: `Около квадрата, периметр которого равен $12$ см, описана окружность. + Найдите длину этой окружности.`, + sol: `Формула периметра квадрата: $P = 4a$.
+Свойство описанной окружности: у окружности, описанной около квадрата, диаметр равен диагонали квадрата, поэтому радиус $R = \\dfrac{a\\sqrt{2}}{2}$, где $a$ — сторона квадрата.
+Формула длины окружности: $C = 2\\pi R$.
+Шаг 1. Находим сторону квадрата.
+$$P = 4a = 12 \\implies a = 3\\text{ см}.$$ +Шаг 2. Находим радиус описанной окружности. +$$R = \\dfrac{a\\sqrt{2}}{2} = \\dfrac{3\\sqrt{2}}{2}\\text{ см}.$$ +Шаг 3. Находим длину окружности. +$$C = 2\\pi R = 2\\pi \\cdot \\dfrac{3\\sqrt{2}}{2} = 3\\pi\\sqrt{2}\\text{ см}.$$ + + + + + + + A + B + C + D + R + O + a = 3 + +
Ответ: $C=3\\pi\\sqrt{2}$ см
` + }, + { + text: `Найдите наибольшее целое значение переменной, при котором сумма дробей + $\\dfrac{2x-1}{5}$ и $\\dfrac{3-4x}{7}$ неотрицательна.`, + sol: `Метод решения линейного неравенства с дробями: приводим к общему знаменателю, домножаем на положительное число (знак не меняется), решаем.
+Шаг 1. «Неотрицательна» значит «не меньше нуля», то есть $\\geq 0$. Записываем: +$$\\dfrac{2x-1}{5} + \\dfrac{3-4x}{7} \\geq 0.$$ +Шаг 2. Приводим к общему знаменателю $35$: +$$\\dfrac{7(2x-1) + 5(3-4x)}{35} \\geq 0.$$ +Шаг 3. Раскрываем скобки в числителе: +$$\\dfrac{14x-7+15-20x}{35} \\geq 0 \\iff \\dfrac{-6x+8}{35} \\geq 0.$$ +Шаг 4. Так как $35\\gt 0$, неравенство равносильно тому, что числитель $\\geq 0$: +$$-6x+8 \\geq 0 \\iff -6x \\geq -8 \\iff x \\leq \\dfrac{4}{3}$$ +(при делении на отрицательное число $-6$ знак меняется).
+Шаг 5. Наибольшее целое $x$, удовлетворяющее $x\\leq \\dfrac{4}{3}\\approx 1{,}33$, — это $x=1$. +
Ответ: $1$
` + }, + { + text: `Упростите выражение + $\\left(\\dfrac{18}{5-\\sqrt{7}} - \\dfrac{44}{7-\\sqrt{5}} - \\dfrac{2}{\\sqrt{7}+\\sqrt{5}}\\right)^{\\!2}$.`, + sol: `Метод рационализации знаменателя: умножаем числитель и знаменатель на сопряжённое выражение, используя формулу $(a-b)(a+b)=a^2-b^2$.
+Шаг 1. Преобразуем первую дробь. Сопряжённое к $5-\\sqrt{7}$ — это $5+\\sqrt{7}$: +$$\\dfrac{18}{5-\\sqrt{7}} = \\dfrac{18(5+\\sqrt{7})}{25-7} = \\dfrac{18(5+\\sqrt{7})}{18} = 5+\\sqrt{7}.$$ +Шаг 2. Преобразуем вторую дробь. Сопряжённое к $7-\\sqrt{5}$ — это $7+\\sqrt{5}$: +$$\\dfrac{44}{7-\\sqrt{5}} = \\dfrac{44(7+\\sqrt{5})}{49-5} = \\dfrac{44(7+\\sqrt{5})}{44} = 7+\\sqrt{5}.$$ +Шаг 3. Преобразуем третью дробь. Сопряжённое к $\\sqrt{7}+\\sqrt{5}$ — это $\\sqrt{7}-\\sqrt{5}$: +$$\\dfrac{2}{\\sqrt{7}+\\sqrt{5}} = \\dfrac{2(\\sqrt{7}-\\sqrt{5})}{7-5} = \\dfrac{2(\\sqrt{7}-\\sqrt{5})}{2} = \\sqrt{7}-\\sqrt{5}.$$ +Шаг 4. Подставляем и приводим подобные: +$$(5+\\sqrt{7})-(7+\\sqrt{5})-(\\sqrt{7}-\\sqrt{5}) = 5+\\sqrt{7}-7-\\sqrt{5}-\\sqrt{7}+\\sqrt{5} = -2.$$ +Шаг 5. Возводим в квадрат: $(-2)^2 = 4$. +
Ответ: $4$
` + }, + { + text: `Дан параллелограмм $ABCD$. На стороне $BC$ взята некоторая точка $M$. + Отрезок $DM$ пересекает диагональ $AC$ в точке $K$. + Площади треугольников $MCK$ и $DCK$ равны соответственно $4$ см² и $6$ см². + Найдите площадь параллелограмма.`, + sol: ` + + + + + + + + A + B + C + D + M + K + 4 + 6 + +Используемые факты: +
    +
  • Треугольники с общей вершиной и основаниями на одной прямой относятся как длины оснований.
    • +
  • Признак подобия по двум углам.
    • +
  • Площадь параллелограмма равна произведению стороны на высоту: $S = a \\cdot h$.
    • +
+Шаг 1. Отношение $MK : KD$.
+Треугольники $MCK$ и $DCK$ имеют общую вершину $C$, а их основания $MK$ и $KD$ лежат на одной прямой $DM$. Значит, их площади относятся как длины оснований: +$$\\dfrac{S_{MCK}}{S_{DCK}} = \\dfrac{MK}{KD} = \\dfrac{4}{6} = \\dfrac{2}{3}.$$ +Шаг 2. Подобие $\\triangle CKM \\sim \\triangle AKD$.
+Так как $BC \\parallel AD$, то $\\angle KCM = \\angle KAD$ (накрест лежащие при секущей $AC$), а $\\angle CKM = \\angle AKD$ (вертикальные). По признаку подобия по двум углам: +$$\\dfrac{CM}{AD} = \\dfrac{KM}{KD} = \\dfrac{2}{3} \\implies CM = \\dfrac{2}{3} AD.$$ +Шаг 3. Площадь $\\triangle CDM$. +$$S_{CDM} = S_{MCK} + S_{DCK} = 4 + 6 = 10\\text{ см}^2.$$ +Шаг 4. Связь с высотой параллелограмма.
+Пусть $h$ — высота параллелограмма, опущенная на $BC$ (она же высота $\\triangle CDM$ из вершины $D$): +$$S_{CDM} = \\dfrac{1}{2} \\cdot CM \\cdot h = \\dfrac{1}{2} \\cdot \\dfrac{2}{3} AD \\cdot h = \\dfrac{1}{3} AD \\cdot h.$$ +Значит, $AD \\cdot h = 3 \\cdot 10 = 30$.
+Шаг 5. Площадь параллелограмма. +$$S_{ABCD} = AD \\cdot h = 30\\text{ см}^2.$$ +
Ответ: $30$ см²
` + }, + { + text: `Решите систему уравнений + $$\\begin{cases} xy - x - y = 5, \\\\[4pt] x^2 + y^2 - x - y = 18. \\end{cases}$$`, + sol: `Тождество: $(x+y)^2 = x^2 + 2xy + y^2$, поэтому $x^2 + y^2 = (x+y)^2 - 2xy$.
+Обратная теорема Виета: если $x + y = s$ и $xy = p$, то $x$ и $y$ — корни уравнения $t^2 - st + p = 0$.
+Шаг 1. Замена $s = x + y$.
+Из первого уравнения: $xy - s = 5 \\implies xy = s + 5$.
+Из второго уравнения: $x^2 + y^2 - s = 18 \\implies x^2 + y^2 = s + 18$.
+Шаг 2. Применяем тождество. +$$x^2 + y^2 = s^2 - 2xy \\implies s + 18 = s^2 - 2(s+5).$$ +Шаг 3. Получаем квадратное уравнение относительно $s$. +$$s + 18 = s^2 - 2s - 10 \\implies s^2 - 3s - 28 = 0.$$ +Дискриминант: $D = 9 + 112 = 121 = 11^2$, корни $s_1 = 7,\\; s_2 = -4$.
+Шаг 4. Случай 1: $s = 7$.
+Тогда $xy = 7 + 5 = 12$. По обратной теореме Виета $x, y$ — корни $t^2 - 7t + 12 = 0$, то есть $t = 3$ или $t = 4$.
+Получаем $(x; y) = (3; 4)$ или $(4; 3)$.
+Проверка: $3 \\cdot 4 - 3 - 4 = 12 - 7 = 5$ ✓; $9 + 16 - 7 = 18$ ✓.
+Шаг 5. Случай 2: $s = -4$.
+Тогда $xy = -4 + 5 = 1$. По обратной теореме Виета $x, y$ — корни $t^2 + 4t + 1 = 0$.
+Дискриминант: $D = 16 - 4 = 12$, корни $t = \\dfrac{-4 \\pm 2\\sqrt{3}}{2} = -2 \\pm \\sqrt{3}$.
+Получаем $(x; y) = (-2 + \\sqrt{3};\\, -2 - \\sqrt{3})$ или $(-2 - \\sqrt{3};\\, -2 + \\sqrt{3})$. +
Ответ: $(3;\\,4),\\ (4;\\,3),\\ (-2+\\sqrt{3};\\,-2-\\sqrt{3}),\\ (-2-\\sqrt{3};\\,-2+\\sqrt{3})$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v59.js b/frontend/js/exam9/variants/v59.js new file mode 100644 index 0000000..0e12b39 --- /dev/null +++ b/frontend/js/exam9/variants/v59.js @@ -0,0 +1,163 @@ +VARIANTS[59] = { + label: "Вариант 59", + tasks: [ + { + text: `Определите рисунок, на котором изображён график функции $y = (x-2)^2$:`, + figure: ``, + sol: `Анализ функции $y=(x-2)^2$:
+ Это парабола, полученная сдвигом графика $y=x^2$ вправо на $2$ единицы.
+
    +
  • Вершина параболы: $(2;\\,0)$
    • +
  • Ветви направлены вверх (коэффициент при $x^2$ равен $1>0$)
    • +
  • Ось симметрии: $x=2$
    • +
  • Пересечение с осью $Oy$: $y(0)=(0-2)^2=4$, точка $(0;\\,4)$
    • +
  • Касается оси $Ox$ в точке $(2;\\,0)$ (двойной корень)
    • +
  • $y \\geq 0$ при всех $x$
    • +
+ Нужно выбрать рисунок, на котором парабола имеет вершину в $(2;\\,0)$ и проходит через $(0;\\,4)$. +
Ответ: парабола с вершиной $(2;\\,0)$, ветви вверх.
` + }, + { + text: `Результат деления многочлена $8m^2 - 16m^3$ на одночлен $2m$ имеет вид:`, + opts: [ + ["а", "$16m^3 - 32m^4$"], ["б", "$4m^2 - 8m^3$"], ["в", "$4m - 8m^3$"], + ["г", "$4m - 8m^2$"], ["д", "$-4m^2$"], + ], + sol: `Делим каждый член многочлена на одночлен $2m$:
+ $\\dfrac{8m^2-16m^3}{2m} = \\dfrac{8m^2}{2m} - \\dfrac{16m^3}{2m} = 4m - 8m^2.$ +
Ответ: г) $4m - 8m^2$.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "диагонали квадрата равны;"], + ["б", "периметр прямоугольника со сторонами $a$ и $b$ равен $P = 2(a+b)$;"], + ["в", "вписанный угол в $2$ раза меньше соответствующего центрального угла;"], + ["г", "$\\sin 45^{\\circ} = 1$?"], + ], + sol: `Проверим утверждения:
+
    +
  • а) Диагонали квадрата равны — верно.
    • +
  • б) $P=2(a+b)$ — верно.
    • +
  • в) Вписанный угол вдвое меньше центрального, опирающегося на ту же дугу — верно.
    • +
  • г) $\\sin 45^{\\circ} = \\dfrac{\\sqrt{2}}{2} \\approx 0{,}707 \\neq 1$ — НЕ верно.
    • +
+
Ответ: г).
` + }, + { + text: `Найдите значение выражения + $15^0 + \\sqrt{16} - \\left(\\dfrac{1}{3}\\right)^{-1} - \\sqrt{\\dfrac{1}{9}}$.`, + sol: `Вычислим каждое слагаемое:
+ $15^0 = 1;\\quad \\sqrt{16}=4;\\quad \\left(\\dfrac{1}{3}\\right)^{-1}=3;\\quad \\sqrt{\\dfrac{1}{9}}=\\dfrac{1}{3}.$
+ Подставим:
+ $1 + 4 - 3 - \\dfrac{1}{3} = 2 - \\dfrac{1}{3} = \\dfrac{6-1}{3} = \\dfrac{5}{3}.$ +
Ответ: $\\dfrac{5}{3} \\approx 1{,}667$.
` + }, + { + text: `Найдите сумму целых решений неравенства $-3 < -2x + 5 \\leq 9$.`, + sol: `Правило: при делении (или умножении) неравенства на отрицательное число знак неравенства меняется на противоположный.
+ Шаг 1. Выписываем неравенство. + $$-3 \\lt -2x + 5 \\leq 9.$$ + Шаг 2. Вычитаем $5$ из всех частей. + $$-3 - 5 \\lt -2x \\leq 9 - 5,$$ + $$-8 \\lt -2x \\leq 4.$$ + Шаг 3. Делим все части на $-2$.
+ Делим на отрицательное число, поэтому оба знака меняются: + $$\\dfrac{-8}{-2} \\gt x \\geq \\dfrac{4}{-2},$$ + $$4 \\gt x \\geq -2 \\iff -2 \\leq x \\lt 4.$$ + Шаг 4. Выписываем целые решения. + $$-2,\\; -1,\\; 0,\\; 1,\\; 2,\\; 3.$$ + Шаг 5. Находим сумму. + $$-2 + (-1) + 0 + 1 + 2 + 3 = 3.$$ +
Ответ: $3$.
` + }, + { + text: `Дан правильный многоугольник со стороной, равной $4$ см. + Сумма всех его внутренних углов равна $1800^{\\circ}$. + Найдите периметр этого многоугольника.`, + sol: `Формула суммы внутренних углов выпуклого $n$-угольника: + $$S_{\\text{углов}} = (n - 2) \\cdot 180^{\\circ}.$$ + Свойство правильного многоугольника: все стороны равны.
+ Шаг 1. Находим число сторон $n$.
+ По условию сумма углов равна $1800^{\\circ}$, поэтому + $$(n - 2) \\cdot 180^{\\circ} = 1800^{\\circ} \\implies n - 2 = 10 \\implies n = 12.$$ + Шаг 2. Находим периметр.
+ Так как многоугольник правильный, все стороны равны $4$ см, значит + $$P = n \\cdot a = 12 \\cdot 4 = 48\\text{ см}.$$ +
Ответ: $48$ см.
` + }, + { + text: `Найдите среднее арифметическое абсцисс точек пересечения графиков функций, + заданных формулами $y = 12 - x - 2x^2$ и $y = 3x^2 - 5x + 3$.`, + sol: `Теорема Виета: для уравнения $ax^2+bx+c=0$ сумма корней равна $-\\dfrac{b}{a}$.
+ Шаг 1. В точках пересечения значения функций равны, поэтому приравниваем правые части: + $$12 - x - 2x^2 = 3x^2 - 5x + 3.$$ + Шаг 2. Переносим всё в одну сторону и приводим подобные: + $$3x^2 - 5x + 3 - 12 + x + 2x^2 = 0 \\implies 5x^2 - 4x - 9 = 0.$$ + Шаг 3. Не решая уравнение, по теореме Виета находим сумму корней: + $$x_1 + x_2 = -\\dfrac{-4}{5} = \\dfrac{4}{5}.$$ + Шаг 4. Среднее арифметическое — это полусумма: + $$\\dfrac{x_1+x_2}{2} = \\dfrac{4/5}{2} = \\dfrac{2}{5} = 0{,}4.$$ +
Ответ: $0{,}4$.
` + }, + { + text: `Смешали $30$-процентный раствор соляной кислоты с $10$-процентным + и получили $600$ г $15$-процентного раствора соляной кислоты. + Сколько граммов каждого раствора было взято?`, + sol: `Пусть $x$ г — масса $30\\%$ раствора, $y$ г — масса $10\\%$ раствора.
+ По условию массы суммируются: $x + y = 600.$
+ Масса чистой кислоты: $0{,}3x + 0{,}1y = 0{,}15 \\cdot 600 = 90.$
+ Получим систему:
+ $\\begin{cases} x+y=600,\\\\ 3x+y=900. \\end{cases}$
+ Вычтем из второго первое: $2x = 300 \\implies x = 150$ г.
+ Тогда $y = 600 - 150 = 450$ г.
+ Проверка: $0{,}3\\cdot 150 + 0{,}1\\cdot 450 = 45+45 = 90$ г кислоты. Верно. +
Ответ: $150$ г ($30\\%$) и $450$ г ($10\\%$).
` + }, + { + text: `Высота прямоугольного треугольника, проведённая к гипотенузе, делит треугольник + на два треугольника, площади которых равны $4$ см² и $16$ см². + Найдите гипотенузу.`, + sol: ` + + + + A + B + C + H + 2 + 8 + h=4 +
+ Пусть $CH=h$ — высота из прямого угла $C$ к гипотенузе $AB$, причём $AH=a$ (меньший отрезок), $HB=b$ (больший отрезок).
+ Площади полученных треугольников:
+ $S_1 = \\dfrac{1}{2}\\cdot a \\cdot h = 4,\\quad S_2 = \\dfrac{1}{2}\\cdot b \\cdot h = 16.$
+ Разделив, получим $\\dfrac{a}{b}=\\dfrac{4}{16}=\\dfrac{1}{4} \\implies b=4a.$
+ По свойству высоты прямоугольного треугольника к гипотенузе: $h^2 = a\\cdot b = a\\cdot 4a = 4a^2 \\implies h = 2a.$
+ Подставим в $S_1$: $\\dfrac{1}{2}\\cdot a \\cdot 2a = a^2 = 4 \\implies a = 2$ см.
+ Тогда $b = 4\\cdot 2 = 8$ см, $h = 4$ см.
+ Гипотенуза: $AB = a + b = 2 + 8 = 10$ см. +
Ответ: $10$ см.
` + }, + { + text: `Упростите выражение + $\\sqrt{x + 2\\sqrt{x-1}} + \\sqrt{x - 2\\sqrt{x-1}}$ при $1 \\leq x \\leq 2$.`, + sol: `Метод выделения полного квадрата и формула $\\sqrt{a^2}=|a|$.
+ Шаг 1. Представим $x$ удобным образом: $x = (x-1) + 1$. Тогда первое подкоренное: + $$x + 2\\sqrt{x-1} = (x-1) + 2\\sqrt{x-1} + 1 = \\left(\\sqrt{x-1}+1\\right)^2$$ + по формуле квадрата суммы $a^2+2ab+b^2=(a+b)^2$ (здесь $a=\\sqrt{x-1}$, $b=1$).
+ Шаг 2. Аналогично для второго: + $$x - 2\\sqrt{x-1} = (x-1) - 2\\sqrt{x-1} + 1 = \\left(\\sqrt{x-1}-1\\right)^2.$$ + Шаг 3. Извлекаем корни, помня что $\\sqrt{a^2}=|a|$: + $$\\sqrt{x+2\\sqrt{x-1}} = \\left|\\sqrt{x-1}+1\\right| = \\sqrt{x-1}+1,$$ + так как $\\sqrt{x-1}+1 \\geq 0$ (модуль не нужен).
+ $$\\sqrt{x-2\\sqrt{x-1}} = \\left|\\sqrt{x-1}-1\\right|.$$ + Шаг 4. Раскрываем второй модуль. По условию $1 \\leq x \\leq 2$, значит $\\sqrt{x-1} \\in [0;\\,1]$, поэтому $\\sqrt{x-1}-1 \\leq 0$, и + $$\\left|\\sqrt{x-1}-1\\right| = 1-\\sqrt{x-1}.$$ + Шаг 5. Складываем: + $$\\left(\\sqrt{x-1}+1\\right) + \\left(1-\\sqrt{x-1}\\right) = 2.$$ +
Ответ: $2$.
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v60.js b/frontend/js/exam9/variants/v60.js new file mode 100644 index 0000000..4cf7293 --- /dev/null +++ b/frontend/js/exam9/variants/v60.js @@ -0,0 +1,138 @@ +VARIANTS[60] = { + label: "Вариант 60", + tasks: [ + { + text: `Определите рисунок, на котором изображён график функции $y = x^2 - 2$:`, + figure: ``, + sol: `Парабола $y=x^2-2$: вершина $(0;-2)$, ветви вверх.
Ответ: парабола с вершиной $(0;-2)$, ветви вверх.
` + }, + { + text: `Результат деления многочлена $10a^3 - 15a^2$ на одночлен $5a$ имеет вид:`, + opts: [ + ["а", "$50a^4 - 75a^3$"], ["б", "$-a^2$"], ["в", "$2a^2 - 3a$"], + ["г", "$2a^2 - 3$"], ["д", "$2a^3 - 3a^2$"], + ], + sol: `$\\dfrac{10a^3-15a^2}{5a}=2a^2-3a$.
Ответ: в) $2a^2-3a$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "диагонали квадрата перпендикулярны;"], + ["б", "периметр параллелограмма со сторонами $a$ и $b$ равен $P = 2a + 2b$;"], + ["в", "$\\cos 45^{\\circ} = 1$;"], + ["г", "центральный угол окружности в $2$ раза больше вписанного угла, опирающегося на ту же дугу?"], + ], + sol: `а) верно; б) $P=2a+2b$ — верно; в) $\\cos45^{\\circ}=\\dfrac{\\sqrt{2}}{2}\\neq1$ — НЕВЕРНО; г) верно.
Ответ: в)
` + }, + { + text: `Найдите значение выражения + $12^0 + \\sqrt{36} - \\left(\\dfrac{1}{2}\\right)^{-1} - \\sqrt{\\dfrac{1}{16}}$.`, + sol: `$1+6-2-\\tfrac{1}{4}=5-\\tfrac{1}{4}=\\dfrac{19}{4}$.
Ответ: $\\dfrac{19}{4}$
` + }, + { + text: `Найдите сумму целых решений неравенства $-7 < -3x + 2 \\leq 5$.`, + sol: `Правило: при делении неравенства на отрицательное число знаки меняются на противоположные.
+ Шаг 1. Выписываем неравенство: + $$-7 \\lt -3x + 2 \\leq 5.$$ + Шаг 2. Вычитаем $2$ из всех частей: + $$-9 \\lt -3x \\leq 3.$$ + Шаг 3. Делим на $-3$ (знаки меняются): + $$3 \\gt x \\geq -1 \\iff -1 \\leq x \\lt 3.$$ + Шаг 4. Целые решения: $-1,\\; 0,\\; 1,\\; 2$.
+ Шаг 5. Сумма: $-1 + 0 + 1 + 2 = 2$. +
Ответ: $2$
` + }, + { + text: `Дан правильный многоугольник с периметром, равным $140$ см. + Сумма всех его внутренних углов равна $900^{\\circ}$. + Найдите длину стороны этого многоугольника.`, + sol: `Формула суммы внутренних углов выпуклого $n$-угольника: + $$S_{\\text{углов}} = (n - 2) \\cdot 180^{\\circ}.$$ + Свойство правильного многоугольника: все стороны равны, значит $P = n \\cdot a$.
+ Шаг 1. Находим число сторон $n$.
+ По условию сумма углов равна $900^{\\circ}$: + $$(n - 2) \\cdot 180^{\\circ} = 900^{\\circ} \\implies n - 2 = 5 \\implies n = 7.$$ + Шаг 2. Находим длину стороны.
+ Периметр $P = n \\cdot a$, откуда + $$a = \\dfrac{P}{n} = \\dfrac{140}{7} = 20\\text{ см}.$$ +
Ответ: $20$ см
` + }, + { + text: `Найдите среднее арифметическое абсцисс точек пересечения графиков функций, + заданных формулами $y = 4x^2 + x$ и $y = 2 - 4x - 3x^2$.`, + sol: `Теорема Виета: для уравнения $ax^2+bx+c=0$ сумма корней равна $-\\dfrac{b}{a}$.
+ Шаг 1. В точках пересечения ординаты совпадают, поэтому приравниваем правые части: + $$4x^2 + x = 2 - 4x - 3x^2.$$ + Шаг 2. Переносим всё в одну сторону и приводим подобные: + $$4x^2 + x - 2 + 4x + 3x^2 = 0 \\implies 7x^2 + 5x - 2 = 0.$$ + Шаг 3. По теореме Виета сумма корней: + $$x_1 + x_2 = -\\dfrac{5}{7}.$$ + Шаг 4. Среднее арифметическое — это полусумма: + $$\\dfrac{x_1+x_2}{2} = -\\dfrac{5}{14}.$$ +
Ответ: $-\\dfrac{5}{14}$
` + }, + { + text: `Имеется лом стали двух сортов с содержанием никеля $5\\%$ и $20\\%$. + Сколько тонн металла каждого сорта надо взять, чтобы получить $150$ т стали + с содержанием никеля $10\\%$?`, + sol: `Метод составления системы уравнений по двум условиям: масса смеси = сумма масс компонентов; масса чистого вещества — тоже сумма по компонентам.
+Шаг 1. Вводим переменные. Пусть $x$ т — масса лома с содержанием никеля $5\\%$, $y$ т — масса лома с содержанием $20\\%$.
+Шаг 2. Составляем первое уравнение (общая масса смеси равна $150$ т): +$$x + y = 150.$$ +Шаг 3. Составляем второе уравнение по массе чистого никеля. В первом ломе никеля $0{,}05x$ т, во втором — $0{,}20y$ т. В готовой смеси никеля $10\\%$ от $150$ т, то есть $15$ т: +$$0{,}05x + 0{,}20y = 15.$$ +Шаг 4. Решаем систему. Умножим второе уравнение на $20$, чтобы избавиться от десятичных: +$$x + 4y = 300.$$ +Вычтем из этого уравнения первое: +$$3y = 150 \\implies y = 50\\text{ т}.$$ +Шаг 5. Находим $x$: +$$x = 150 - 50 = 100\\text{ т}.$$ +Шаг 6. Проверка: масса никеля $0{,}05\\cdot 100 + 0{,}20\\cdot 50 = 5 + 10 = 15$ т — совпадает с условием. +
Ответ: $100$ т ($5\\%$) и $50$ т ($20\\%$).
` + }, + { + text: `Высота прямоугольного треугольника, проведённая к гипотенузе, делит треугольник + на два треугольника, площади которых равны $6$ см² и $54$ см². + Найдите гипотенузу.`, + sol: ` + + + + A + B + C + H + 2 + 18 + h=6 +
+Пусть $CH = h$ — высота из прямого угла $C$ к гипотенузе $AB$, $AH = a$, $HB = b$.
+Площади треугольников: +$$S_1 = \\dfrac{1}{2}\\cdot a\\cdot h = 6, \\quad S_2 = \\dfrac{1}{2}\\cdot b\\cdot h = 54.$$ +Делим $S_2$ на $S_1$: $\\dfrac{b}{a} = \\dfrac{54}{6} = 9 \\implies b = 9a.$
+По свойству высоты прямоугольного треугольника: $h^2 = a\\cdot b = 9a^2 \\implies h = 3a.$
+Подставим в $S_1$: $\\dfrac{1}{2}\\cdot a\\cdot 3a = \\dfrac{3a^2}{2} = 6 \\implies a^2 = 4 \\implies a = 2$ см.
+Тогда $b = 9\\cdot 2 = 18$ см.
+Гипотенуза: $AB = a + b = 2 + 18 = 20$ см. +
Ответ: $20$ см.
` + }, + { + text: `Упростите выражение + $\\sqrt{x + 6\\sqrt{x-9}} + \\sqrt{x - 6\\sqrt{x-9}}$ при $x > 18$.`, + sol: `Метод выделения полного квадрата и формула $\\sqrt{a^2}=|a|$.
+ Шаг 1. Представляем $x$ удобным образом: $x = (x-9) + 9$. Тогда первое подкоренное выражение раскладывается по формуле квадрата суммы $a^2+2ab+b^2=(a+b)^2$ (здесь $a=\\sqrt{x-9}$, $b=3$): + $$x + 6\\sqrt{x-9} = (x-9) + 2\\cdot\\sqrt{x-9}\\cdot 3 + 9 = \\left(\\sqrt{x-9}+3\\right)^2.$$ + Шаг 2. Аналогично для второго подкоренного (квадрат разности): + $$x - 6\\sqrt{x-9} = \\left(\\sqrt{x-9}-3\\right)^2.$$ + Шаг 3. Извлекаем корни по правилу $\\sqrt{a^2}=|a|$: + $$\\sqrt{x+6\\sqrt{x-9}} = \\left|\\sqrt{x-9}+3\\right| = \\sqrt{x-9}+3,$$ + так как $\\sqrt{x-9}+3 \\gt 0$ (модуль не нужен).
+ $$\\sqrt{x-6\\sqrt{x-9}} = \\left|\\sqrt{x-9}-3\\right|.$$ + Шаг 4. Раскрываем второй модуль. По условию $x \\gt 18$, значит $x-9 \\gt 9$ и $\\sqrt{x-9} \\gt 3$, поэтому $\\sqrt{x-9}-3 \\gt 0$ и + $$\\left|\\sqrt{x-9}-3\\right| = \\sqrt{x-9}-3.$$ + Шаг 5. Складываем результаты: + $$\\left(\\sqrt{x-9}+3\\right) + \\left(\\sqrt{x-9}-3\\right) = 2\\sqrt{x-9}.$$ +
Ответ: $2\\sqrt{x-9}$.
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v61.js b/frontend/js/exam9/variants/v61.js new file mode 100644 index 0000000..fb936ea --- /dev/null +++ b/frontend/js/exam9/variants/v61.js @@ -0,0 +1,184 @@ +VARIANTS[61] = { + label: "Вариант 61", + tasks: [ + { + text: `Определите, какое из данных уравнений является приведённым:`, + opts: [ + ["а", "$4x^2 + x = 0$"], ["б", "$2 - 4x - 3x^2 = 0$"], ["в", "$x^2 - 3x + 2 = 0$"], + ["г", "$3x + 2 = 0$"], ["д", "$-x^2 - 3x + 4 = 0$"], + ], + sol: `Приведённое квадратное уравнение — это уравнение вида $x^2 + px + q = 0$, у которого коэффициент при $x^2$ равен $1$.
+ Проверим варианты: +
    +
  • а) $4x^2 + x = 0$ — коэффициент при $x^2$ равен $4$;
    • +
  • б) $2 - 4x - 3x^2 = 0$ — коэффициент при $x^2$ равен $-3$;
    • +
  • в) $x^2 - 3x + 2 = 0$ — коэффициент при $x^2$ равен $1$ — приведённое;
    • +
  • г) $3x + 2 = 0$ — линейное, не квадратное;
    • +
  • д) $-x^2 - 3x + 4 = 0$ — коэффициент при $x^2$ равен $-1$.
    • +
+
Ответ: в) $x^2 - 3x + 2 = 0$.
` + }, + { + text: `Какое из данных выражений равно выражению $\\dfrac{\\sqrt{16}}{2}$:`, + opts: [ + ["а", "$\\sqrt{8}$"], ["б", "$8$"], ["в", "$\\sqrt{2}$"], ["г", "$2$"], ["д", "$4$"], + ], + sol: `Вычислим: $\\sqrt{16} = 4$, тогда + $$\\dfrac{\\sqrt{16}}{2} = \\dfrac{4}{2} = 2.$$ +
Ответ: г) $2$.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "острый угол больше $0^{\\circ}$ и меньше $90^{\\circ}$;"], + ["б", "если $\\alpha$ — острый угол, то $\\sin^2\\alpha + \\cos^2\\alpha = 1$;"], + ["в", "центр окружности, описанной около прямоугольного треугольника, лежит на середине гипотенузы;"], + ["г", "в любом параллелограмме все углы равны между собой?"], + ], + sol: `Разберём утверждения: +
    +
  • а) определение острого угла — верно;
    • +
  • б) основное тригонометрическое тождество — верно;
    • +
  • в) свойство описанной около прямоугольного треугольника окружности — верно;
    • +
  • г) в произвольном параллелограмме противолежащие углы равны, но соседние углы в общем случае различны (равенство всех углов выполняется только в прямоугольнике) — не верно.
    • +
+
Ответ: г).
` + }, + { + text: `Найдите наименьшее целое решение неравенства $\\dfrac{5}{x-1} \\geq 0$.`, + sol: `Числитель $5 > 0$, поэтому знак дроби совпадает со знаком знаменателя.
+ Дробь определена при $x \\ne 1$. Условие $\\dfrac{5}{x-1} \\geq 0$ выполняется, когда + $$x - 1 > 0 \\;\\Longleftrightarrow\\; x > 1.$$ + Целые числа, удовлетворяющие неравенству $x > 1$: $2,\\;3,\\;4,\\;\\ldots$
+ Наименьшее из них — $2$. +
Ответ: $2$.
` + }, + { + text: `Сократите дробь $\\dfrac{4x^2 - 9y^2}{2x - 3y}$ и найдите её значение, если $x = 0{,}5$, $y = \\dfrac{2}{3}$.`, + sol: `Формула разности квадратов: $a^2 - b^2 = (a - b)(a + b)$.
+ Шаг 1. Раскладываем числитель.
+ Заметим, что $4x^2 = (2x)^2$ и $9y^2 = (3y)^2$, значит числитель — разность квадратов: + $$4x^2 - 9y^2 = (2x)^2 - (3y)^2 = (2x - 3y)(2x + 3y).$$ + Шаг 2. Сокращаем дробь.
+ В числителе и знаменателе есть общий множитель $(2x - 3y)$: + $$\\dfrac{(2x - 3y)(2x + 3y)}{2x - 3y} = 2x + 3y, \\quad 2x \\neq 3y.$$ + Шаг 3. Подставляем значения $x = 0{,}5$, $y = \\dfrac{2}{3}$. + $$2 \\cdot 0{,}5 + 3 \\cdot \\dfrac{2}{3} = 1 + 2 = 3.$$ +
Ответ: $2x + 3y$; значение равно $3$.
` + }, + { + text: `В параллелограмме $ABCD$ углы $BAC$ и $DAC$ равны $45^{\\circ}$ и $30^{\\circ}$ соответственно, $AB = 6$ см. Найдите длину стороны $BC$.`, + sol: ` + + + + + + + + A + B + C + D + 45° + 30° + 6 + + Теорема синусов: в треугольнике стороны пропорциональны синусам противолежащих углов: + $$\\dfrac{a}{\\sin A} = \\dfrac{b}{\\sin B} = \\dfrac{c}{\\sin C}.$$ + Свойство параллельных прямых: при пересечении секущей накрест лежащие углы равны.
+ Шаг 1. Находим $\\angle BCA$.
+ В параллелограмме $BC \\parallel AD$, а $AC$ — секущая. Углы $\\angle BCA$ и $\\angle DAC$ накрест лежащие, значит + $$\\angle BCA = \\angle DAC = 30^{\\circ}.$$ + Шаг 2. Записываем известное в $\\triangle ABC$.
+ $\\angle BAC = 45^{\\circ}$, $\\angle BCA = 30^{\\circ}$, $AB = 6$ — сторона, противолежащая углу $\\angle BCA$.
+ Сторона $BC$ противолежит углу $\\angle BAC$.
+ Шаг 3. Применяем теорему синусов. + $$\\dfrac{AB}{\\sin\\angle BCA} = \\dfrac{BC}{\\sin\\angle BAC},$$ + $$\\dfrac{6}{\\sin 30^{\\circ}} = \\dfrac{BC}{\\sin 45^{\\circ}}.$$ + Шаг 4. Подставляем табличные значения $\\sin 30^{\\circ} = \\dfrac{1}{2}$ и $\\sin 45^{\\circ} = \\dfrac{\\sqrt{2}}{2}$. + $$BC = \\dfrac{6 \\sin 45^{\\circ}}{\\sin 30^{\\circ}} = \\dfrac{6 \\cdot \\tfrac{\\sqrt{2}}{2}}{\\tfrac{1}{2}} = 6\\sqrt{2}\\text{ см}.$$ +
Ответ: $BC = 6\\sqrt{2}$ см.
` + }, + { + text: `График функции $f(x) = kx + b$ изображён на рисунке. + Используя график функции, найдите $k$ и $b$. + Запишите формулу функции $y = f(x)$.`, + figure: ``, + sol: `Метод (по графику): +
    +
  • Свободный коэффициент $b$ — это ордината точки пересечения графика с осью $Oy$ (значение $y$ при $x = 0$).
    • +
  • Угловой коэффициент $k$ находится по двум точкам $(x_1;\\,y_1)$ и $(x_2;\\,y_2)$ графика по формуле + $$k = \\dfrac{y_2 - y_1}{x_2 - x_1}.$$
    • +
  • После нахождения $k$ и $b$ записываем формулу $y = kx + b$.
    • +
+ Например, если на графике видно, что прямая пересекает ось $Oy$ в точке $(0;\\,b)$ и проходит через точку $(x_1;\\,y_1)$, то + $$k = \\dfrac{y_1 - b}{x_1 - 0},\\qquad f(x) = kx + b.$$ +
Ответ: $f(x) = kx + b$, где $k$ и $b$ определяются по графику указанным способом.
` + }, + { + text: `В геометрической прогрессии произведение третьего и десятого членов равно $120$. + Чему равно произведение одиннадцатого и второго членов этой прогрессии?`, + sol: `Формула $n$-го члена геометрической прогрессии: + $$b_n = b_1 \\cdot q^{n-1},$$ + где $b_1$ — первый член, $q$ — знаменатель прогрессии.
+ Шаг 1. Записываем произведение двух членов с номерами $p$ и $q$. + $$b_p \\cdot b_q = b_1 q^{p-1} \\cdot b_1 q^{q-1} = b_1^2 \\cdot q^{p+q-2}.$$ + Шаг 2. Делаем вывод о произведении.
+ Произведение зависит только от суммы номеров $p + q$. Значит, если $p + q = r + s$, то + $$b_p \\cdot b_q = b_r \\cdot b_s.$$ + Шаг 3. Сравниваем суммы номеров.
+ Для пары $(3, 10)$: сумма $3 + 10 = 13$.
+ Для пары $(2, 11)$: сумма $2 + 11 = 13$.
+ Суммы равны, значит и произведения равны: + $$b_2 \\cdot b_{11} = b_3 \\cdot b_{10} = 120.$$ +
Ответ: $120$.
` + }, + { + text: `На изготовление комплекта деталей для холодильной установки бригада затратила $\\dfrac{2}{5}$ часа + и выпустила за $8$-часовую смену $640$ деталей. Сколько деталей выпустит бригада за смену, + если время на изготовление комплекта деталей будет равно $\\dfrac{4}{15}$ часа?`, + sol: `Метод решения задачи по действиям: постепенно находим число комплектов, число деталей в одном комплекте, а затем общее количество деталей.
+ Шаг 1. Находим, сколько комплектов бригада выпускала за смену в первом случае. Делим всё время смены на время одного комплекта (по правилу деления на дробь — умножаем на обратную): + $$8 : \\dfrac{2}{5} = 8 \\cdot \\dfrac{5}{2} = 20\\text{ комплектов}.$$ + Шаг 2. Находим, сколько деталей в одном комплекте. По условию за смену выпущено $640$ деталей, всего $20$ комплектов: + $$640 : 20 = 32\\text{ детали в комплекте}.$$ + Шаг 3. Находим, сколько комплектов будет выпущено за смену при новой норме времени $\\dfrac{4}{15}$ часа: + $$8 : \\dfrac{4}{15} = 8 \\cdot \\dfrac{15}{4} = 30\\text{ комплектов}.$$ + Шаг 4. Так как в каждом комплекте по $32$ детали, общее количество деталей: + $$30 \\cdot 32 = 960\\text{ деталей}.$$ +
Ответ: $960$ деталей.
` + }, + { + text: `В угол $A$ вписана окружность с радиусом $6$ см и центром в точке $O_1$. + Расстояние от центра этой окружности до вершины угла равно $30$ см. + Найдите радиус меньшей окружности с центром в точке $O_2$, + которая касается сторон данного угла и данной окружности.`, + figure: ``, + sol: ` + + + + + + O₁ + R = 6 + + + O₂ + r + A + + Центры обеих окружностей лежат на биссектрисе угла $A$. Пусть $\\angle A = 2\\alpha$.
+ Из прямоугольного треугольника, образованного вершиной $A$, центром $O_1$ и точкой касания окружности со стороной угла: + $$\\sin\\alpha = \\dfrac{R}{AO_1} = \\dfrac{6}{30} = \\dfrac{1}{5}.$$ + Аналогично для меньшей окружности радиуса $r$ с центром $O_2$: + $$\\sin\\alpha = \\dfrac{r}{AO_2} \\;\\Longrightarrow\\; AO_2 = \\dfrac{r}{\\sin\\alpha} = 5r.$$ + Окружности касаются внешним образом, $O_2$ лежит между $A$ и $O_1$, поэтому + $$O_1 O_2 = R + r,\\qquad O_1 O_2 = AO_1 - AO_2 = 30 - 5r.$$ + Получаем уравнение: + $$30 - 5r = 6 + r \\;\\Longrightarrow\\; 6r = 24 \\;\\Longrightarrow\\; r = 4\\;\\text{см}.$$ +
Ответ: $r = 4$ см.
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v62.js b/frontend/js/exam9/variants/v62.js new file mode 100644 index 0000000..683dd92 --- /dev/null +++ b/frontend/js/exam9/variants/v62.js @@ -0,0 +1,176 @@ +VARIANTS[62] = { + label: "Вариант 62", + tasks: [ + { // V62_MARKER + text: `Определите, какое из данных уравнений является приведённым:`, + opts: [ + ["а", "$2x^2 + x = 0$"], ["б", "$3 - 4x - 2x^2 = 0$"], ["в", "$x^2 - 4x - 12 = 0$"], + ["г", "$3x - 7 = 0$"], ["д", "$-x^2 - 5x + 6 = 0$"], + ], + sol: `Приведённое квадратное уравнение — это уравнение вида $x^2 + px + q = 0$, у которого коэффициент при $x^2$ равен $1$.
+ Проверим варианты: +
    +
  • а) $2x^2 + x = 0$ — коэффициент при $x^2$ равен $2$;
    • +
  • б) $3 - 4x - 2x^2 = 0$ — коэффициент при $x^2$ равен $-2$;
    • +
  • в) $x^2 - 4x - 12 = 0$ — коэффициент при $x^2$ равен $1$ — приведённое;
    • +
  • г) $3x - 7 = 0$ — линейное, не квадратное;
    • +
  • д) $-x^2 - 5x + 6 = 0$ — коэффициент при $x^2$ равен $-1$.
    • +
+
Ответ: в) $x^2 - 4x - 12 = 0$.
` + }, + { + text: `Какое из данных выражений равно выражению $\\dfrac{\\sqrt{36}}{2}$:`, + opts: [ + ["а", "$\\sqrt{3}$"], ["б", "$2\\sqrt{3}$"], ["в", "$\\dfrac{\\sqrt{6}}{2}$"], ["г", "$3$"], ["д", "$3\\sqrt{2}$"], + ], + sol: `Вычислим: $\\sqrt{36} = 6$, тогда + $$\\dfrac{\\sqrt{36}}{2} = \\dfrac{6}{2} = 3.$$ +
Ответ: г) $3$.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "тупой угол больше $90^{\\circ}$ и меньше $180^{\\circ}$;"], + ["б", "если $\\alpha$ — острый угол, то $\\operatorname{tg}\\alpha \\cdot \\operatorname{ctg}\\alpha = 1$;"], + ["в", "радиус окружности, описанной около прямоугольного треугольника, равен половине гипотенузы;"], + ["г", "в любом параллелограмме все стороны равны между собой?"], + ], + sol: `Разберём утверждения: +
    +
  • а) определение тупого угла — верно;
    • +
  • б) $\\operatorname{tg}\\alpha \\cdot \\operatorname{ctg}\\alpha = \\dfrac{\\sin\\alpha}{\\cos\\alpha}\\cdot\\dfrac{\\cos\\alpha}{\\sin\\alpha} = 1$ — верно;
    • +
  • в) свойство описанной около прямоугольного треугольника окружности — верно;
    • +
  • г) в произвольном параллелограмме противолежащие стороны равны, но соседние стороны в общем случае различны (все стороны равны только в ромбе) — не верно.
    • +
+
Ответ: г).
` + }, + { + text: `Найдите наибольшее целое решение неравенства $\\dfrac{5}{x+1} \\leq 0$.`, + sol: `Числитель $5 > 0$, поэтому знак дроби совпадает со знаком знаменателя.
+ Дробь определена при $x \\ne -1$. Условие $\\dfrac{5}{x+1} \\leq 0$ выполняется, когда + $$x + 1 < 0 \\;\\Longleftrightarrow\\; x < -1.$$ + Целые числа, удовлетворяющие неравенству $x < -1$: $\\ldots,\\;-4,\\;-3,\\;-2$.
+ Наибольшее из них — $-2$. +
Ответ: $-2$.
` + }, + { + text: `Сократите дробь $\\dfrac{x^2 - 16y^2}{x - 4y}$ и найдите её значение, если $x = 1$, $y = \\dfrac{1}{2}$.`, + sol: `Формула разности квадратов: $a^2 - b^2 = (a - b)(a + b)$.
+ Шаг 1. Раскладываем числитель.
+ Так как $16y^2 = (4y)^2$, числитель — разность квадратов: + $$x^2 - 16y^2 = x^2 - (4y)^2 = (x - 4y)(x + 4y).$$ + Шаг 2. Сокращаем дробь на общий множитель $(x - 4y)$. + $$\\dfrac{(x - 4y)(x + 4y)}{x - 4y} = x + 4y, \\quad x \\neq 4y.$$ + Шаг 3. Подставляем $x = 1$, $y = \\dfrac{1}{2}$. + $$1 + 4 \\cdot \\dfrac{1}{2} = 1 + 2 = 3.$$ +
Ответ: $x + 4y$; значение равно $3$.
` + }, + { + text: `В параллелограмме $ABCD$ углы $BAC$ и $DAC$ равны $30^{\\circ}$ и $45^{\\circ}$ соответственно, $AD = 8$ см. Найдите длину стороны $AB$.`, + sol: ` + + + A + B + C + D + 30° + 45° + 8 + + Теорема синусов: стороны треугольника пропорциональны синусам противолежащих углов.
+ Свойство параллелограмма: противоположные стороны равны и параллельны.
+ Шаг 1. Находим $\\angle BCA$.
+ $BC \\parallel AD$, $AC$ — секущая. Накрест лежащие углы равны: + $$\\angle BCA = \\angle DAC = 45^{\\circ}.$$ + Шаг 2. Записываем известное в $\\triangle ABC$.
+ $\\angle BAC = 30^{\\circ}$, $\\angle BCA = 45^{\\circ}$. Так как $BC = AD = 8$ см (противоположные стороны параллелограмма равны), сторона $BC = 8$ напротив угла $\\angle BAC$.
+ Искомая сторона $AB$ — напротив $\\angle BCA$.
+ Шаг 3. По теореме синусов. + $$\\dfrac{AB}{\\sin\\angle BCA} = \\dfrac{BC}{\\sin\\angle BAC},$$ + $$\\dfrac{AB}{\\sin 45^{\\circ}} = \\dfrac{8}{\\sin 30^{\\circ}}.$$ + Шаг 4. Подставляем $\\sin 30^{\\circ} = \\dfrac{1}{2}$, $\\sin 45^{\\circ} = \\dfrac{\\sqrt{2}}{2}$. + $$AB = \\dfrac{8 \\sin 45^{\\circ}}{\\sin 30^{\\circ}} = \\dfrac{8 \\cdot \\tfrac{\\sqrt{2}}{2}}{\\tfrac{1}{2}} = 8\\sqrt{2}\\text{ см}.$$ +
Ответ: $AB = 8\\sqrt{2}$ см.
` + }, + { + text: `График функции $f(x) = kx + b$ изображён на рисунке. + Используя график функции, найдите $k$ и $b$. + Запишите формулу функции $y = f(x)$.`, + figure: ``, + sol: `Метод (по графику): +
    +
  • Свободный коэффициент $b$ — это ордината точки пересечения графика с осью $Oy$ (значение $y$ при $x = 0$).
    • +
  • Угловой коэффициент $k$ находится по двум точкам $(x_1;\\,y_1)$ и $(x_2;\\,y_2)$ графика по формуле + $$k = \\dfrac{y_2 - y_1}{x_2 - x_1}.$$
    • +
  • После нахождения $k$ и $b$ записываем формулу $y = kx + b$.
    • +
+ Например, если на графике видно, что прямая пересекает ось $Oy$ в точке $(0;\\,b)$ и проходит через точку $(x_1;\\,y_1)$, то + $$k = \\dfrac{y_1 - b}{x_1 - 0},\\qquad f(x) = kx + b.$$ +
Ответ: $f(x) = kx + b$, где $k$ и $b$ определяются по графику указанным способом.
` + }, + { + text: `В геометрической прогрессии произведение четвёртого и двенадцатого членов равно $200$. + Чему равно произведение второго и четырнадцатого членов этой прогрессии?`, + sol: `Формула $n$-го члена геометрической прогрессии: + $$b_n = b_1 \\cdot q^{n-1}.$$ + Шаг 1. Записываем произведение двух членов. + $$b_p \\cdot b_q = b_1 q^{p-1} \\cdot b_1 q^{q-1} = b_1^2 \\cdot q^{p+q-2}.$$ + Шаг 2. Свойство.
+ Произведение зависит только от суммы номеров. Значит, если $p + q = r + s$, то + $$b_p \\cdot b_q = b_r \\cdot b_s.$$ + Шаг 3. Сравниваем суммы.
+ Для $(4, 12)$: $4 + 12 = 16$.
+ Для $(2, 14)$: $2 + 14 = 16$.
+ Суммы совпадают, поэтому + $$b_2 \\cdot b_{14} = b_4 \\cdot b_{12} = 200.$$ +
Ответ: $200$.
` + }, + { + text: `На изготовление комплекта деталей для автопогрузчика бригада затратила $\\dfrac{1}{4}$ часа + и выпустила за $8$-часовую смену $480$ деталей. Сколько деталей выпустит бригада за смену, + если время на изготовление комплекта деталей будет равно $\\dfrac{4}{17}$ часа?`, + sol: `Метод решения задачи по действиям: находим число комплектов за смену, число деталей в одном комплекте, а затем общее количество деталей при новой норме.
+ Шаг 1. Находим, сколько комплектов выпускалось за смену в первом случае. Делим время смены на время одного комплекта (по правилу деления на дробь — умножаем на обратную): + $$8 : \\dfrac{1}{4} = 8 \\cdot 4 = 32\\text{ комплекта}.$$ + Шаг 2. Находим, сколько деталей в одном комплекте. По условию за смену выпущено $480$ деталей, всего $32$ комплекта: + $$480 : 32 = 15\\text{ деталей в комплекте}.$$ + Шаг 3. Находим, сколько комплектов будет выпущено за смену при новой норме $\\dfrac{4}{17}$ часа: + $$8 : \\dfrac{4}{17} = 8 \\cdot \\dfrac{17}{4} = 34\\text{ комплекта}.$$ + Шаг 4. Так как в каждом комплекте по $15$ деталей, общее количество деталей за смену: + $$34 \\cdot 15 = 510\\text{ деталей}.$$ +
Ответ: $510$ деталей.
` + }, + { + text: `В угол $A$ вписана окружность с радиусом $8$ см и центром в точке $O_1$. + Расстояние от центра этой окружности до вершины угла равно $40$ см. + Найдите радиус большей окружности с центром в точке $O_2$, + которая касается сторон данного угла и данной окружности.`, + figure: ``, + sol: ` + + + + + + O₁ + R=8 + + + O₂ + r + A + + Центры обеих окружностей лежат на биссектрисе угла $A$. Пусть $\\angle A = 2\\alpha$.
+ Из прямоугольного треугольника, образованного вершиной $A$, центром $O_1$ и точкой касания: + $$\\sin\\alpha = \\dfrac{R}{AO_1} = \\dfrac{8}{40} = \\dfrac{1}{5}.$$ + Для большей окружности радиуса $r$ с центром $O_2$ (лежит дальше от $A$, чем $O_1$): + $$\\sin\\alpha = \\dfrac{r}{AO_2} \\;\\Longrightarrow\\; AO_2 = 5r.$$ + Окружности касаются внешним образом, $O_1$ лежит между $A$ и $O_2$, поэтому + $$O_1 O_2 = AO_2 - AO_1 = 5r - 40,\\qquad O_1 O_2 = R + r = 8 + r.$$ + Получаем уравнение: + $$5r - 40 = 8 + r \\;\\Longrightarrow\\; 4r = 48 \\;\\Longrightarrow\\; r = 12\\;\\text{см}.$$ +
Ответ: $r = 12$ см.
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v63.js b/frontend/js/exam9/variants/v63.js new file mode 100644 index 0000000..440336b --- /dev/null +++ b/frontend/js/exam9/variants/v63.js @@ -0,0 +1,176 @@ +VARIANTS[63] = { + label: "Вариант 63", + tasks: [ + { + text: `Определите, какое из данных равенств является верным:`, + opts: [ + ["а", "$a^5 = 5a$"], ["б", "$a^5 = 5a^5$"], ["в", "$a^5 = 5 + a^5$"], + ["г", "$a^5 = a^5$ (т.е. $a \\cdot a \\cdot a \\cdot a \\cdot a = a^5$)"], ["д", "$a^5 = 5 : a^5$"], + ], + sol: `По определению степени с натуральным показателем: + $$a^5 = \\underbrace{a\\cdot a\\cdot a\\cdot a\\cdot a}_{5\\text{ раз}}.$$ + Остальные равенства неверны: $5a$, $5a^5$, $5+a^5$, $5:a^5$ — это другие выражения. +
Ответ: г.
` + }, + { + text: `Частное дробей $\\dfrac{4}{5}$ и $\\dfrac{24}{25}$ равно:`, + opts: [ + ["а", "$\\dfrac{6}{5}$"], ["б", "$\\dfrac{125}{96}$"], ["в", "$\\dfrac{5}{6}$"], + ["г", "$1{,}2$"], ["д", "$\\dfrac{96}{125}$"], + ], + sol: `Деление дробей — это умножение на обратную: + $$\\dfrac{4}{5} : \\dfrac{24}{25} = \\dfrac{4}{5}\\cdot\\dfrac{25}{24} = \\dfrac{4\\cdot 25}{5\\cdot 24} = \\dfrac{100}{120} = \\dfrac{5}{6}.$$ +
Ответ: в.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "если две окружности касаются внешним образом, то сумма их радиусов равна расстоянию между их центрами;"], + ["б", "$\\sin 60^{\\circ} = \\dfrac{\\sqrt{3}}{2}$;"], + ["в", "на плоскости две прямые, перпендикулярные третьей, параллельны между собой;"], + ["г", "у любого ромба все углы прямые?"], + ], + sol: `Проверяем каждое утверждение: +
    +
  • а) верно — это свойство внешнего касания окружностей;
    • +
  • б) верно — табличное значение $\\sin 60^{\\circ} = \\dfrac{\\sqrt{3}}{2}$;
    • +
  • в) верно — признак параллельности прямых на плоскости;
    • +
  • г) неверно — прямые углы есть только у квадрата (частного случая ромба), а не у любого ромба.
    • +
+
Ответ: г.
` + }, + { + text: `Найдите значение выражения $\\left(3\\dfrac{1}{3}\\right)^{-2}$. + В ответ запишите противоположное ему число.`, + sol: `Превратим смешанное число в обыкновенную дробь: $3\\dfrac{1}{3} = \\dfrac{10}{3}$.
+ По свойству $a^{-n} = \\dfrac{1}{a^{n}}$: + $$\\left(\\dfrac{10}{3}\\right)^{-2} = \\left(\\dfrac{3}{10}\\right)^{2} = \\dfrac{9}{100} = 0{,}09.$$ + Противоположное число: $-0{,}09$. +
Ответ: $-0{,}09$ (или $-\\dfrac{9}{100}$).
` + }, + { + text: `Вершина угла $ABC$ лежит на окружности с центром в точке $O$, + а стороны пересекают окружность в точках $A$ и $C$. + Угол $ABO$ равен $40^{\\circ}$, угол $ACO$ равен $30^{\\circ}$. + Найдите величину угла $BOC$.`, + sol: ` + + + O + + + B + + A + + C + + + + + + + + $OA=OB=OC=R$ (радиусы), значит треугольники $OAB$, $OAC$, $OBC$ равнобедренные, и углы при их основаниях равны.
+ В $\\triangle OAB$: $\\angle OAB = \\angle OBA = 40^{\\circ}$.
+ В $\\triangle OAC$: $\\angle OAC = \\angle OCA = 30^{\\circ}$.
+ Тогда $\\angle BAC = \\angle OAB + \\angle OAC = 40^{\\circ} + 30^{\\circ} = 70^{\\circ}$.
+ По сумме углов $\\triangle ABC$: + $$\\angle ABC + \\angle ACB = 180^{\\circ} - 70^{\\circ} = 110^{\\circ}.$$ + Заметим, что $\\angle ABC = 40^{\\circ} + \\angle OBC$ и $\\angle ACB = 30^{\\circ} + \\angle OCB$. Подставляем: + $$40^{\\circ} + 30^{\\circ} + \\angle OBC + \\angle OCB = 110^{\\circ} \\implies \\angle OBC + \\angle OCB = 40^{\\circ}.$$ + В $\\triangle OBC$ ($OB=OC$) углы при основании равны: $\\angle OBC = \\angle OCB = 20^{\\circ}$.
+ Значит $\\angle BOC = 180^{\\circ} - 20^{\\circ} - 20^{\\circ} = 140^{\\circ}$. +
Ответ: $\\angle BOC = 140^{\\circ}$.
` + }, + { + text: `Найдите число, $37\\%$ которого равны значению выражения $4{,}5 : 9 + 3{,}2$.`, + sol: `Правило нахождения числа по его проценту: если $p\\%$ числа $N$ равны $A$, то $N = \\dfrac{A}{p/100}$.
+ Шаг 1. Сначала найдём значение выражения. По порядку действий деление выполняется раньше сложения: + $$4{,}5 : 9 + 3{,}2 = 0{,}5 + 3{,}2 = 3{,}7.$$ + Шаг 2. Обозначим искомое число $N$. По условию $37\\%$ от $N$ равны $3{,}7$, то есть + $$0{,}37\\,N = 3{,}7.$$ + Шаг 3. Находим $N$, разделив обе части на $0{,}37$: + $$N = \\dfrac{3{,}7}{0{,}37} = 10.$$ +
Ответ: $10$.
` + }, + { + text: `График линейной функции проходит через точки $A(-2;\\;-4)$ и $B(0;\\;0)$. + Запишите формулу, задающую эту функцию, + и найдите значение выражения $f(-1) + f(3)$.`, + sol: `Линейная функция имеет вид $f(x) = kx + b$, где $k$ — угловой коэффициент, $b$ — ордината точки пересечения с осью $Oy$ (значение при $x=0$).
+ Шаг 1. Так как точка $B(0;\\,0)$ принадлежит графику, то $f(0) = b = 0$. Значит формула имеет вид + $$f(x) = kx.$$ + Шаг 2. Точка $A(-2;\\,-4)$ тоже принадлежит графику, значит $f(-2) = -4$. Подставляем: + $$-4 = k \\cdot (-2) \\implies k = 2.$$ + Шаг 3. Записываем формулу: $f(x) = 2x$.
+ Шаг 4. Находим значение выражения $f(-1) + f(3)$: + $$f(-1) + f(3) = 2 \\cdot (-1) + 2 \\cdot 3 = -2 + 6 = 4.$$ +
Ответ: $f(x) = 2x$, $\\ f(-1)+f(3) = 4$.
` + }, + { + text: `Решите систему уравнений + $$\\begin{cases} x - 2y = 1, \\\\[4pt] 3x + 4y = 23 \\end{cases}$$ + и найдите разность найденных значений $x$ и $y$.`, + sol: `Метод сложения: уравниваем коэффициенты при одной из переменных так, чтобы при сложении уравнений эта переменная пропала.
+ Шаг 1. Уравниваем коэффициенты при $y$.
+ В первом уравнении коэффициент при $y$ равен $-2$, во втором — $4$. Умножим первое уравнение на $2$, чтобы получить $-4y$: + $$\\begin{cases} 2x - 4y = 2, \\\\ 3x + 4y = 23. \\end{cases}$$ + Шаг 2. Складываем уравнения.
+ $y$ взаимно уничтожается: + $$5x = 25 \\implies x = 5.$$ + Шаг 3. Находим $y$.
+ Подставляем $x = 5$ в первое исходное уравнение $x - 2y = 1$: + $$5 - 2y = 1 \\implies 2y = 4 \\implies y = 2.$$ + Шаг 4. Находим разность. + $$x - y = 5 - 2 = 3.$$ +
Ответ: $x = 5,\\ y = 2,\\ x - y = 3$.
` + }, + { + text: `Собственная скорость катера равна $24$ км/ч. + Через сколько минут катер, двигаясь навстречу плоту, встретит его, + если он находится от плота на расстоянии $12$ км?`, + sol: `Плот плывёт со скоростью течения $v_p$ (км/ч). Катер идёт ему навстречу — значит против течения, его скорость относительно берега $24 - v_p$.
+ Скорость сближения катера и плота: + $$(24 - v_p) + v_p = 24 \\text{ км/ч}.$$ + Скорость течения сокращается, поэтому ответ от неё не зависит.
+ Время до встречи: + $$t = \\dfrac{12}{24} = 0{,}5\\text{ ч} = 30\\text{ мин}.$$ +
Ответ: через $30$ минут.
` + }, + { + text: `Известно, что в равнобедренном треугольнике $ABC$ $AB = BC = 4$. + Найдите $AC$, если медиана $AM = 3$.`, + sol: ` + + + + + + A + + C + + B + + M + AM = 3 + AB = 4 + BC = 4 + + $M$ — середина $BC$, поэтому $BM = MC = \\dfrac{BC}{2} = 2$ см.
+ Идея: используем теорему косинусов дважды с одним и тем же углом $B$ (общим для $\\triangle ABM$ и $\\triangle ABC$). +

+ Шаг 1. Применим теорему косинусов к $\\triangle ABM$ (стороны $AB=4$, $BM=2$, $AM=3$): + $$AM^2 = AB^2 + BM^2 - 2\\cdot AB\\cdot BM\\cdot\\cos\\angle B$$ + $$9 = 16 + 4 - 16\\cos\\angle B$$ + $$16\\cos\\angle B = 11 \\implies \\cos\\angle B = \\dfrac{11}{16}$$ + Шаг 2. Теперь применим теорему косинусов к $\\triangle ABC$ (тот же угол $B$, стороны $AB=BC=4$): + $$AC^2 = AB^2 + BC^2 - 2\\cdot AB\\cdot BC\\cdot\\cos\\angle B$$ + $$AC^2 = 16 + 16 - 2\\cdot4\\cdot4\\cdot\\dfrac{11}{16}$$ + $$AC^2 = 32 - 22 = 10$$ + $$AC = \\sqrt{10}\\text{ см}$$ +
Ответ: $AC = \\sqrt{10}$ см
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v64.js b/frontend/js/exam9/variants/v64.js new file mode 100644 index 0000000..9da5b93 --- /dev/null +++ b/frontend/js/exam9/variants/v64.js @@ -0,0 +1,181 @@ +VARIANTS[64] = { + label: "Вариант 64", + tasks: [ + { + text: `Определите, какое из данных равенств является верным:`, + opts: [ + ["а", "$b^6 = 6b$"], ["б", "$b^6 = 6b^6$"], ["в", "$b^6 = 6 + b^6$"], + ["г", "$b \\cdot b \\cdot b \\cdot b \\cdot b \\cdot b = b^6$"], ["д", "$b^6 = 6 \\cdot b^6$"], + ], + sol: `По определению степени с натуральным показателем: + $$b^6 = \\underbrace{b\\cdot b\\cdot b\\cdot b\\cdot b\\cdot b}_{6\\text{ раз}}.$$ + Остальные равенства неверны: $6b$, $6b^6$, $6+b^6$, $6\\cdot b^6$ — это другие выражения. +
Ответ: г.
` + }, + { + text: `Произведение дробей $\\dfrac{14}{15}$ и $\\dfrac{25}{49}$ равно:`, + opts: [ + ["а", "$\\dfrac{10}{11}$"], ["б", "$\\dfrac{7}{10}$"], ["в", "$\\dfrac{10}{21}$"], + ["г", "$1{,}1$"], ["д", "$2{,}1$"], + ], + sol: `Умножим дроби: + $$\\dfrac{14}{15} \\cdot \\dfrac{25}{49} = \\dfrac{14 \\cdot 25}{15 \\cdot 49} = \\dfrac{350}{735} = \\dfrac{10}{21}.$$ +
Ответ: в) $\\dfrac{10}{21}$.
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "если две окружности касаются, то они имеют единственную общую точку;"], + ["б", "$\\cos 60^{\\circ} = \\dfrac{1}{2}$;"], + ["в", "на плоскости две прямые, параллельные третьей, параллельны между собой;"], + ["г", "у любого прямоугольника все стороны равны?"], + ], + sol: `Проверяем каждое утверждение: +
    +
  • а) верно — определение касания окружностей (внутреннего или внешнего);
    • +
  • б) верно — табличное значение $\\cos 60^{\\circ} = \\dfrac{1}{2}$;
    • +
  • в) верно — транзитивность параллельности на плоскости;
    • +
  • г) неверно — в прямоугольнике все углы прямые, но стороны в общем случае различны (равные стороны у квадрата, являющегося частным случаем прямоугольника).
    • +
+
Ответ: г.
` + }, + { + text: `Найдите значение выражения $\\left(2\\dfrac{3}{4}\\right)^{-2}$. + В ответ запишите противоположное ему число.`, + sol: `Превратим смешанное число в обыкновенную дробь: $2\\dfrac{3}{4} = \\dfrac{11}{4}$.
+ По свойству $a^{-n} = \\dfrac{1}{a^{n}}$: + $$\\left(\\dfrac{11}{4}\\right)^{-2} = \\left(\\dfrac{4}{11}\\right)^{2} = \\dfrac{16}{121}.$$ + Противоположное число: $-\\dfrac{16}{121}$. +
Ответ: $-\\dfrac{16}{121}$.
` + }, + { + text: `Вершина угла $ABC$ лежит на окружности с центром в точке $O$, + а стороны пересекают окружность в точках $A$ и $C$. + Угол $ABO$ равен $20^{\\circ}$, угол $ACO$ равен $40^{\\circ}$. + Найдите величину угла $BOC$.`, + sol: ` + + + O + + B + + A + + C + + + + + + + + $OA=OB=OC=R$ (радиусы), значит треугольники $OAB$, $OAC$, $OBC$ равнобедренные, и углы при их основаниях равны.
+ В $\\triangle OAB$: $\\angle OAB = \\angle OBA = 20^{\\circ}$.
+ В $\\triangle OAC$: $\\angle OAC = \\angle OCA = 40^{\\circ}$.
+ Тогда $\\angle BAC = \\angle OAB + \\angle OAC = 20^{\\circ} + 40^{\\circ} = 60^{\\circ}$.
+ По сумме углов $\\triangle ABC$: + $$\\angle ABC + \\angle ACB = 180^{\\circ} - 60^{\\circ} = 120^{\\circ}.$$ + Заметим, что $\\angle ABC = 20^{\\circ} + \\angle OBC$ и $\\angle ACB = 40^{\\circ} + \\angle OCB$. Подставляем: + $$20^{\\circ} + 40^{\\circ} + \\angle OBC + \\angle OCB = 120^{\\circ} \\implies \\angle OBC + \\angle OCB = 60^{\\circ}.$$ + В $\\triangle OBC$ ($OB=OC$) углы при основании равны: $\\angle OBC = \\angle OCB = 30^{\\circ}$.
+ Значит $\\angle BOC = 180^{\\circ} - 30^{\\circ} - 30^{\\circ} = 120^{\\circ}$. +
Ответ: $\\angle BOC = 120^{\\circ}$.
` + }, + { + text: `Найдите число, $24\\%$ которого равны значению выражения $4{,}5 : 3 + 3{,}3$.`, + sol: `Правило нахождения числа по его проценту: если $p\\%$ числа $N$ равны $A$, то $N=\\dfrac{A}{p/100}$.
+ Шаг 1. Сначала находим значение выражения. По порядку действий сначала выполняется деление, потом сложение: + $$4{,}5 : 3 + 3{,}3 = 1{,}5 + 3{,}3 = 4{,}8.$$ + Шаг 2. Обозначим искомое число $N$. По условию $24\\%$ от $N$ равны $4{,}8$: + $$0{,}24\\,N = 4{,}8.$$ + Шаг 3. Делим обе части на $0{,}24$: + $$N = \\dfrac{4{,}8}{0{,}24} = 20.$$ +
Ответ: $20$.
` + }, + { + text: `График линейной функции проходит через точки $A(3;\\;6)$ и $B(0;\\;0)$. + Запишите формулу, задающую эту функцию, + и найдите значение выражения $f(1) + f(-2)$.`, + sol: `Линейная функция имеет вид $f(x) = kx + b$, где $k$ — угловой коэффициент, $b$ — ордината точки пересечения с осью $Oy$ (значение функции при $x=0$).
+ Шаг 1. Так как точка $B(0;\\,0)$ принадлежит графику, то $f(0)=b=0$. Значит формула имеет вид + $$f(x) = kx.$$ + Шаг 2. Точка $A(3;\\,6)$ тоже принадлежит графику, поэтому $f(3)=6$. Подставляем: + $$6 = k \\cdot 3 \\implies k = 2.$$ + Шаг 3. Записываем формулу: $f(x) = 2x$.
+ Шаг 4. Находим значение выражения $f(1) + f(-2)$: + $$f(1) + f(-2) = 2\\cdot 1 + 2\\cdot(-2) = 2 - 4 = -2.$$ +
Ответ: $f(x) = 2x$, $\\ f(1)+f(-2) = -2$.
` + }, + { + text: `Решите систему уравнений + $$\\begin{cases} 11x - 8y = -53, \\\\[4pt] 9x + 4y = -17 \\end{cases}$$ + и найдите разность найденных значений $x$ и $y$.`, + sol: `Метод сложения: уравниваем коэффициенты при одной из переменных так, чтобы они стали противоположными, а при сложении уравнений она исчезла.
+ Шаг 1. Уравниваем коэффициенты при $y$.
+ В первом уравнении $-8y$, во втором $4y$. Умножим второе уравнение на $2$, чтобы получить $8y$: + $$\\begin{cases} 11x - 8y = -53, \\\\ 18x + 8y = -34. \\end{cases}$$ + Шаг 2. Складываем уравнения.
+ $y$ уничтожается: + $$29x = -87 \\implies x = -3.$$ + Шаг 3. Находим $y$.
+ Подставим $x = -3$ во второе исходное уравнение $9x + 4y = -17$: + $$9 \\cdot (-3) + 4y = -17 \\implies -27 + 4y = -17 \\implies 4y = 10 \\implies y = 2{,}5.$$ + Шаг 4. Находим разность. + $$x - y = -3 - 2{,}5 = -5{,}5.$$ +
Ответ: $x = -3,\\ y = 2{,}5,\\ x - y = -5{,}5$.
` + }, + { + text: `Собственная скорость катера равна $28$ км/ч. + Через сколько минут катер, двигаясь по течению, догонит плот, + если он находится от плота на расстоянии $14$ км?`, + sol: `Плот плывёт со скоростью течения $v_p$ (км/ч). Катер идёт по течению, его скорость относительно берега $28 + v_p$.
+ Скорость сближения катера и плота: + $$(28 + v_p) - v_p = 28 \\text{ км/ч}.$$ + Скорость течения сокращается, поэтому ответ от неё не зависит.
+ Время до встречи: + $$t = \\dfrac{14}{28} = 0{,}5\\text{ ч} = 30\\text{ мин}.$$ +
Ответ: через $30$ минут.
` + }, + { + text: `Известно, что в равнобедренном треугольнике $ABC$ $AB = BC = 6$. + Найдите $AC$, если медиана $AM = 4$.`, + sol: ` + + + + A + + C + + B + + M + AM = 4 + AB = 6 + BC = 6 + + Теорема косинусов: для любого треугольника со сторонами $a, b, c$ и углом $\\gamma$ между сторонами $a$ и $b$: + $$c^2 = a^2 + b^2 - 2ab\\cos\\gamma.$$ + Определение медианы: медиана $AM$ соединяет вершину $A$ с серединой $M$ противоположной стороны $BC$.
+ Идея: в $\\triangle ABM$ и $\\triangle ABC$ есть общий угол $B$. Найдём $\\cos\\angle B$ из первого треугольника, а затем используем его для второго.
+ Шаг 1. Находим $BM$.
+ $M$ — середина $BC$, поэтому + $$BM = MC = \\dfrac{BC}{2} = \\dfrac{6}{2} = 3.$$ + Шаг 2. Применяем теорему косинусов к $\\triangle ABM$.
+ Стороны $AB = 6$, $BM = 3$, $AM = 4$, угол между $AB$ и $BM$ — это $\\angle B$: + $$AM^2 = AB^2 + BM^2 - 2 \\cdot AB \\cdot BM \\cdot \\cos\\angle B,$$ + $$16 = 36 + 9 - 2 \\cdot 6 \\cdot 3 \\cdot \\cos\\angle B,$$ + $$16 = 45 - 36\\cos\\angle B.$$ + Выражаем $\\cos\\angle B$: + $$36\\cos\\angle B = 29 \\implies \\cos\\angle B = \\dfrac{29}{36}.$$ + Шаг 3. Применяем теорему косинусов к $\\triangle ABC$.
+ Стороны $AB = BC = 6$, угол между ними — тот же $\\angle B$: + $$AC^2 = AB^2 + BC^2 - 2 \\cdot AB \\cdot BC \\cdot \\cos\\angle B,$$ + $$AC^2 = 36 + 36 - 2 \\cdot 6 \\cdot 6 \\cdot \\dfrac{29}{36} = 72 - 58 = 14.$$ + Шаг 4. Находим $AC$. + $$AC = \\sqrt{14}\\text{ см}.$$ +
Ответ: $AC = \\sqrt{14}$ см.
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v65.js b/frontend/js/exam9/variants/v65.js new file mode 100644 index 0000000..2abb10e --- /dev/null +++ b/frontend/js/exam9/variants/v65.js @@ -0,0 +1,178 @@ +VARIANTS[65] = { + label: "Вариант 65", + tasks: [ + { + text: `Из данных равенств выберите тождество:`, + opts: [ + ["а", "$x + x + x = x^3$"], ["б", "$x \\cdot x \\cdot x = 3x$"], ["в", "$x + x + x = 3x$"], + ["г", "$x \\cdot x \\cdot x \\cdot x = 4 + x$"], ["д", "$x + x + x = 3 + x$"], + ], + sol: `Тождество — равенство, верное при любых значениях переменной. +
    +
  • а) $x+x+x=x^3$ — неверно: слева $3x$, справа $x^3$;
    • +
  • б) $x\\cdot x\\cdot x=3x$ — неверно: слева $x^3$;
    • +
  • в) $x+x+x=3x$ — верно ✓ (сумма трёх одинаковых слагаемых равна утроенному слагаемому);
    • +
  • г) $x\\cdot x\\cdot x\\cdot x=4+x$ — неверно: слева $x^4$;
    • +
  • д) $x+x+x=3+x$ — неверно: слева $3x$, а $3x\\neq 3+x$ (например, при $x=2$: $6\\neq 5$).
    • +
+
Ответ: в) $x+x+x=3x$
` + }, + { + text: `Определите, в какой из данных точек график функции $y = 2x + 3$ пересекает ось ординат:`, + opts: [ + ["а", "$A(1{,}5;\\;0)$"], ["б", "$B(0;\\;1{,}5)$"], ["в", "$C(3;\\;0)$"], + ["г", "$D(-1{,}5;\\;0)$"], ["д", "$E(0;\\;3)$"], + ], + sol: `Ось ординат ($Oy$) — это прямая $x=0$. Подставим $x=0$ в уравнение функции: +$$y = 2\\cdot 0 + 3 = 3.$$ +Значит, график пересекает ось $Oy$ в точке $(0;\\;3)$. +
Ответ: д) $E(0;\\;3)$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "для сторон треугольника $ABC$ верно $\\dfrac{AB}{\\sin C} = \\dfrac{AC}{\\sin B}$;"], + ["б", "$\\sin 120^{\\circ} = -\\dfrac{\\sqrt{3}}{2}$;"], + ["в", "около прямоугольника всегда можно описать окружность;"], + ["г", "длина окружности находится по формуле $C = 2\\pi R$?"], + ], + sol: `
    +
  • а) Теорема синусов — верно;
    • +
  • б) По формуле приведения: $\\sin 120^{\\circ}=\\sin(180^{\\circ}-60^{\\circ})=\\sin 60^{\\circ}=\\dfrac{\\sqrt{3}}{2}$ (положительное число!) — НЕВЕРНО;
    • +
  • в) Около любого прямоугольника описывается окружность (центр — точка пересечения диагоналей) — верно;
    • +
  • г) Формула длины окружности $C=2\\pi R$ — верно.
    • +
+
Ответ: б)
` + }, + { + text: `Какая из следующих последовательностей является геометрической прогрессией? Ответ обоснуйте.
+ а) $5;\\; 15;\\; 45;\\; \\ldots$  + б) $5;\\; 10;\\; 15;\\; \\ldots$  + в) $1;\\; 4;\\; 9;\\; 16;\\; \\ldots$  + г) $\\dfrac{1}{2};\\; \\dfrac{1}{3};\\; \\dfrac{1}{4};\\; \\dfrac{1}{5};\\; \\ldots$`, + sol: `Геометрическая прогрессия — последовательность, в которой каждый член (начиная со второго) получается умножением предыдущего на одно и то же число $q$ (знаменатель прогрессии). +
    +
  • а) $5;\\; 15;\\; 45;\\;\\ldots$   $\\dfrac{15}{5}=3,\\;\\dfrac{45}{15}=3$ — отношение постоянное, $q=3$. Это ГП ✓
    • +
  • б) $5;\\; 10;\\; 15;\\;\\ldots$   $\\dfrac{10}{5}=2,\\;\\dfrac{15}{10}=1{,}5$ — отношения разные. Это арифметическая прогрессия ($d=5$).
    • +
  • в) $1;\\; 4;\\; 9;\\; 16;\\;\\ldots$   $\\dfrac{4}{1}=4,\\;\\dfrac{9}{4}=2{,}25$ — отношения разные (квадраты натуральных).
    • +
  • г) $\\dfrac{1}{2};\\;\\dfrac{1}{3};\\;\\dfrac{1}{4};\\;\\ldots$   $\\dfrac{1/3}{1/2}=\\dfrac{2}{3},\\;\\dfrac{1/4}{1/3}=\\dfrac{3}{4}$ — отношения разные.
    • +
+
Ответ: а) $5;\\; 15;\\; 45;\\;\\ldots$ — ГП со знаменателем $q=3$.
` + }, + { + text: `Упростите выражение $\\dfrac{m^3}{m+1} \\cdot \\dfrac{m^2+2m+1}{2m^4}$.`, + sol: `Формула квадрата суммы: $a^2 + 2ab + b^2 = (a+b)^2$. +
Правило умножения дробей: $\\dfrac{a}{b}\\cdot\\dfrac{c}{d}=\\dfrac{a\\cdot c}{b\\cdot d}$. +
Шаг 1. Найдём ОДЗ. Знаменатели не должны равняться нулю: $m+1\\neq 0$ и $2m^4\\neq 0$, значит $m\\neq -1$ и $m\\neq 0$. +
Шаг 2. Разложим числитель второй дроби по формуле квадрата суммы. Замечаем, что $m^2+2m+1 = m^2 + 2\\cdot m\\cdot 1 + 1^2 = (m+1)^2$: +$$\\dfrac{m^3}{m+1}\\cdot\\dfrac{(m+1)^2}{2m^4}.$$ +Шаг 3. Перемножим дроби и сократим общие множители. В числителе появляется $m^3(m+1)^2$, в знаменателе — $(m+1)\\cdot 2m^4$. Сокращаем $(m+1)$ в первой степени и $m^3$ из степени $m^4$: +$$\\dfrac{m^3(m+1)^2}{(m+1)\\cdot 2m^4} = \\dfrac{m+1}{2m}.$$ +
Ответ: $\\dfrac{m+1}{2m}$.
` + }, + { + text: `Около окружности с радиусом $4$ см описана равнобедренная трапеция, + площадь которой равна $80$ см². Найдите длину боковой стороны этой трапеции.`, + sol: `Свойство 1. Высота трапеции, описанной около окружности, равна диаметру вписанной окружности: +$$h = 2r = 2\\cdot 4 = 8\\text{ см}.$$ +Свойство 2. Если четырёхугольник описан около окружности, то суммы его противоположных сторон равны. Для равнобедренной трапеции с основаниями $a,\\;b$ и боковыми сторонами $c$: +$$a+b = 2c.$$ +Из формулы площади трапеции $S=\\dfrac{a+b}{2}\\cdot h$: +$$80 = \\dfrac{a+b}{2}\\cdot 8 \\implies a+b = 20\\text{ см}.$$ + + + + + + + + + + + + + A + D + B + C + + a + b + c + c + + h=8 + r=4 + +Тогда $2c = a+b = 20\\implies c = 10$ см. +
Ответ: $c = 10$ см.
` + }, + { + text: `Сравните корень уравнения $\\dfrac{4}{5}\\left(\\dfrac{6}{25}x - 1\\right) = 4$ + с числом $\\left(\\dfrac{1}{5}\\right)^{-2}$.`, + sol: `Свойство степени с отрицательным показателем: $\\left(\\dfrac{a}{b}\\right)^{-n} = \\left(\\dfrac{b}{a}\\right)^{n}$. +
Шаг 1. Решим уравнение. Сначала избавимся от множителя $\\dfrac{4}{5}$ перед скобкой — разделим обе части на $\\dfrac{4}{5}$, то есть умножим на $\\dfrac{5}{4}$: +$$\\dfrac{6}{25}x - 1 = 4\\cdot\\dfrac{5}{4} = 5.$$ +Шаг 2. Переносим $-1$ в правую часть (меняем знак): +$$\\dfrac{6}{25}x = 5 + 1 = 6.$$ +Шаг 3. Чтобы найти $x$, умножим обе части на $\\dfrac{25}{6}$ (число, обратное к $\\dfrac{6}{25}$): +$$x = 6\\cdot\\dfrac{25}{6} = 25.$$ +Шаг 4. Вычислим число для сравнения. По свойству степени: +$$\\left(\\dfrac{1}{5}\\right)^{-2} = 5^{2} = 25.$$ +Шаг 5. Сравниваем: $x = 25$ и $25$. Значит, корень уравнения равен числу $\\left(\\dfrac{1}{5}\\right)^{-2}$. +
Ответ: корень уравнения равен числу $\\left(\\dfrac{1}{5}\\right)^{-2}$ (оба равны $25$).
` + }, + { + text: `Найдите сумму целых значений аргумента, для которых график функции + $y = \\dfrac{2x-10}{x^2+x-12}$ расположен выше прямой $y = 1$.`, + sol: `Условие: $\\dfrac{2x-10}{x^2+x-12} > 1.$ Перенесём всё в одну часть: +$$\\dfrac{2x-10}{x^2+x-12} - 1 > 0 \\iff \\dfrac{2x-10-(x^2+x-12)}{x^2+x-12} > 0 \\iff \\dfrac{-x^2+x+2}{x^2+x-12} > 0.$$ +Умножим числитель и знаменатель на $-1$ (знак неравенства меняется): +$$\\dfrac{x^2-x-2}{x^2+x-12} \\lt 0.$$ +Разложим: $x^2-x-2=(x-2)(x+1)$, $\\;x^2+x-12=(x+4)(x-3)$: +$$\\dfrac{(x-2)(x+1)}{(x+4)(x-3)} \\lt 0.$$ +Корни: $-4,\\;-1,\\;2,\\;3$ (точки $-4$ и $3$ не входят — ОДЗ). +
Метод интервалов: + + + +
интервал$x\\lt-4$$(-4;-1)$$(-1;2)$$(2;3)$$x\\gt 3$
знак дроби$+$$-$$+$$-$$+$
+Решение: $x\\in(-4;\\;-1)\\cup(2;\\;3)$. +
Целые значения: в $(-4;-1)$ — это $-3,\\;-2$; в $(2;3)$ — целых нет. +
Сумма: $-3+(-2)=-5$. +
Ответ: $-5$.
` + }, + { + text: `Дана окружность, длина которой равна $12\\pi$. + Найдите площадь сектора круга, ограниченного этой окружностью, + если угол этого сектора равен $40^{\\circ}$.`, + sol: `Формула длины окружности: $C = 2\\pi R$. +
Формула площади сектора с центральным углом $\\alpha^{\\circ}$: $S_{\\text{сект}} = \\dfrac{\\alpha}{360^{\\circ}}\\cdot \\pi R^{2}$. +
Шаг 1. Найдём радиус. По условию длина окружности равна $12\\pi$, значит: +$$2\\pi R = 12\\pi \\implies R = 6\\text{ см}.$$ +Шаг 2. Подставим в формулу площади сектора $\\alpha = 40^{\\circ}$ и $R = 6$: +$$S_{\\text{сект}} = \\dfrac{40}{360}\\cdot \\pi\\cdot 6^{2} = \\dfrac{1}{9}\\cdot 36\\pi = 4\\pi.$$ +
Ответ: $4\\pi$ (кв. ед.).
` + }, + { + text: `На соревнованиях управляемых планеров первый планер пролетел на $20\\%$, + или на $1080$ м, меньше второго. Скорость первого планера на $20\\%$, + или на $2$ м/с, больше скорости второго. + Сколько минут находился в воздухе каждый планер?`, + sol: `Связь процентов и десятичной дроби: $20\\% = \\dfrac{20}{100} = 0{,}2$. +
Формула пути: $S = v\\cdot t$, откуда $t = \\dfrac{S}{v}$. +
Шаг 1. Найдём путь второго планера. По условию $20\\%$ от $S_{2}$ — это $1080$ м, так как разница $S_{2} - S_{1}$ одновременно есть и $20\\%$ от $S_{2}$, и $1080$ м. Составим уравнение: +$$0{,}2\\cdot S_{2} = 1080 \\implies S_{2} = \\dfrac{1080}{0{,}2} = 5400\\text{ м}.$$ +Тогда путь первого планера: +$$S_{1} = S_{2} - 1080 = 5400 - 1080 = 4320\\text{ м}.$$ +Шаг 2. Найдём скорость второго планера. Аналогично, $20\\%$ от $v_{2}$ равны $2$ м/с: +$$0{,}2\\cdot v_{2} = 2 \\implies v_{2} = \\dfrac{2}{0{,}2} = 10\\text{ м/с}.$$ +Скорость первого планера больше на $2$ м/с: +$$v_{1} = v_{2} + 2 = 12\\text{ м/с}.$$ +Шаг 3. Найдём время полёта каждого планера по формуле $t = \\dfrac{S}{v}$ и переведём секунды в минуты ($60$ с $= 1$ мин): +$$t_{1} = \\dfrac{S_{1}}{v_{1}} = \\dfrac{4320}{12} = 360\\text{ с} = 6\\text{ мин};$$ +$$t_{2} = \\dfrac{S_{2}}{v_{2}} = \\dfrac{5400}{10} = 540\\text{ с} = 9\\text{ мин}.$$ +
Ответ: 1-й планер — $6$ мин, 2-й планер — $9$ мин.
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v66.js b/frontend/js/exam9/variants/v66.js new file mode 100644 index 0000000..6122263 --- /dev/null +++ b/frontend/js/exam9/variants/v66.js @@ -0,0 +1,170 @@ +VARIANTS[66] = { + label: "Вариант 66", + tasks: [ + { + text: `Из данных равенств выберите тождество:`, + opts: [ + ["а", "$y + y + y + y + y = 5 + y$"], ["б", "$y \\cdot y \\cdot y \\cdot y \\cdot y \\cdot y = 6y^6$"], ["в", "$y + y + y = y^2$"], + ["г", "$y + y + y + y = y^4$"], ["д", "$y \\cdot y \\cdot y \\cdot y = y^4$"], + ], + sol: `Тождество — равенство, верное при любых значениях переменной. +
    +
  • а) $y+y+y+y+y=5+y$ — неверно: слева $5y$, справа $5+y$ (не одно и то же);
    • +
  • б) $y\\cdot y\\cdot y\\cdot y\\cdot y\\cdot y=6y^6$ — неверно: слева $y^6$, не $6y^6$;
    • +
  • в) $y+y+y=y^2$ — неверно: слева $3y$, справа $y^2$ (например, при $y=2$: $6\\neq 4$);
    • +
  • г) $y+y+y+y=y^4$ — неверно: слева $4y$, справа $y^4$;
    • +
  • д) $y\\cdot y\\cdot y\\cdot y=y^4$ — верно ✓ (произведение четырёх одинаковых множителей равно четвёртой степени).
    • +
+
Ответ: д) $y\\cdot y\\cdot y\\cdot y=y^4$
` + }, + { + text: `Определите, в какой из данных точек график функции $y = 2x + 5$ пересекает ось ординат:`, + opts: [ + ["а", "$A(2{,}5;\\;0)$"], ["б", "$B(0;\\;2{,}5)$"], ["в", "$C(5;\\;0)$"], + ["г", "$D(-2{,}5;\\;0)$"], ["д", "$E(0;\\;5)$"], + ], + sol: `Ось ординат ($Oy$) — это прямая $x=0$. Подставим $x=0$ в уравнение функции: +$$y = 2\\cdot 0 + 5 = 5.$$ +Значит, график пересекает ось $Oy$ в точке $(0;\\;5)$. +
Ответ: д) $E(0;\\;5)$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "в ромб всегда можно вписать окружность;"], + ["б", "для сторон треугольника $ABC$ верно $\\dfrac{AC}{\\sin B} = \\dfrac{BC}{\\sin A}$;"], + ["в", "$\\cos 120^{\\circ} = \\dfrac{1}{2}$;"], + ["г", "площадь круга находится по формуле $S = \\pi R^2$?"], + ], + sol: `
    +
  • а) В ромб всегда можно вписать окружность (сумма противоположных сторон равна) — верно;
    • +
  • б) Теорема синусов — верно;
    • +
  • в) $\\cos 120^{\\circ}=\\cos(180^{\\circ}-60^{\\circ})=-\\cos 60^{\\circ}=-\\dfrac{1}{2}$ (отрицательное число!) — НЕВЕРНО;
    • +
  • г) Формула площади круга $S=\\pi R^2$ — верно.
    • +
+
Ответ: в)
` + }, + { + text: `Какая из следующих последовательностей является геометрической прогрессией? Ответ обоснуйте.
+ а) $25;\\; 35;\\; 45;\\; \\ldots$  + б) $0{,}2;\\; 0{,}02;\\; 0{,}002;\\; \\ldots$  + в) $\\dfrac{4}{3};\\; 9;\\; \\dfrac{5}{16};\\; \\ldots$  + г) $5;\\; -5;\\; -15;\\; \\ldots$`, + sol: `Геометрическая прогрессия — последовательность, в которой каждый член (начиная со второго) получается умножением предыдущего на одно и то же число $q$ (знаменатель прогрессии). +
    +
  • а) $25;\\; 35;\\; 45;\\;\\ldots$   $\\dfrac{35}{25}=1{,}4,\\;\\dfrac{45}{35}\\approx1{,}286$ — отношения разные (АП с $d=10$). Не ГП.
    • +
  • б) $0{,}2;\\; 0{,}02;\\; 0{,}002;\\;\\ldots$   $\\dfrac{0{,}02}{0{,}2}=0{,}1,\\;\\dfrac{0{,}002}{0{,}02}=0{,}1$ — отношение постоянное, $q=0{,}1$. Это ГП ✓
    • +
  • в) $\\dfrac{4}{3};\\; 9;\\; \\dfrac{5}{16};\\;\\ldots$   $\\dfrac{9}{4/3}=\\dfrac{27}{4},\\;\\dfrac{5/16}{9}=\\dfrac{5}{144}$ — отношения разные. Не ГП.
    • +
  • г) $5;\\; -5;\\; -15;\\;\\ldots$   $\\dfrac{-5}{5}=-1,\\;\\dfrac{-15}{-5}=3$ — отношения разные (АП с $d=-10$). Не ГП.
    • +
+
Ответ: б) $0{,}2;\\; 0{,}02;\\; 0{,}002;\\;\\ldots$ — ГП со знаменателем $q=0{,}1$.
` + }, + { + text: `Упростите выражение $\\dfrac{m^2}{m-1} \\cdot \\dfrac{m^2-2m+1}{2m^3}$.`, + sol: `Формула квадрата разности: $a^2 - 2ab + b^2 = (a-b)^2$. +
Правило умножения дробей: $\\dfrac{a}{b}\\cdot\\dfrac{c}{d}=\\dfrac{ac}{bd}$. +
Шаг 1. Найдём ОДЗ. Знаменатели не равны нулю: $m-1\\neq 0$ и $2m^3\\neq 0$, значит $m\\neq 1$ и $m\\neq 0$. +
Шаг 2. Разложим числитель второй дроби по формуле квадрата разности: $m^2-2m+1 = m^2 - 2\\cdot m\\cdot 1 + 1^2 = (m-1)^2$: +$$\\dfrac{m^2}{m-1}\\cdot\\dfrac{(m-1)^2}{2m^3}.$$ +Шаг 3. Перемножим дроби и сократим общие множители $(m-1)$ и $m^2$: +$$\\dfrac{m^2(m-1)^2}{(m-1)\\cdot 2m^3} = \\dfrac{m-1}{2m}.$$ +
Ответ: $\\dfrac{m-1}{2m}$.
` + }, + { + text: `Около окружности с радиусом $3$ см описана равнобедренная трапеция, + площадь которой равна $24$ см². Найдите длину боковой стороны этой трапеции.`, + sol: `Свойство 1. Высота трапеции, описанной около окружности, равна диаметру вписанной окружности: +$$h = 2r = 2\\cdot 3 = 6\\text{ см}.$$ +Свойство 2. Для равнобедренной трапеции, описанной около окружности, суммы противоположных сторон равны: +$$a+b = 2c.$$ +Из формулы площади трапеции $S=\\dfrac{a+b}{2}\\cdot h$: +$$24 = \\dfrac{a+b}{2}\\cdot 6 \\implies a+b = 8\\text{ см}.$$ + + + + + + + A + D + B + C + a + b + c + c + h=6 + r=3 + +Тогда $2c = a+b = 8\\implies c = 4$ см. +
Ответ: $c = 4$ см.
` + }, + { + text: `Сравните корень уравнения $\\dfrac{4}{3}\\left(\\dfrac{1}{2}x - 1\\right) = 4$ + с числом $\\left(\\dfrac{1}{2}\\right)^{-3}$.`, + sol: `Свойство степени с отрицательным показателем: $\\left(\\dfrac{a}{b}\\right)^{-n} = \\left(\\dfrac{b}{a}\\right)^{n}$. +
Шаг 1. Решим уравнение. Сначала избавимся от множителя $\\dfrac{4}{3}$ перед скобкой — разделим обе части на $\\dfrac{4}{3}$, то есть умножим на $\\dfrac{3}{4}$: +$$\\dfrac{1}{2}x - 1 = 4\\cdot\\dfrac{3}{4} = 3.$$ +Шаг 2. Переносим $-1$ в правую часть: +$$\\dfrac{1}{2}x = 3 + 1 = 4.$$ +Шаг 3. Умножим обе части на $2$: +$$x = 4\\cdot 2 = 8.$$ +Шаг 4. Вычислим число для сравнения по свойству степени: +$$\\left(\\dfrac{1}{2}\\right)^{-3} = 2^{3} = 8.$$ +Шаг 5. Сравниваем: $x = 8$ и $8$. Значит, корень уравнения равен числу $\\left(\\dfrac{1}{2}\\right)^{-3}$. +
Ответ: корень уравнения равен числу $\\left(\\dfrac{1}{2}\\right)^{-3}$ (оба равны $8$).
` + }, + { + text: `Найдите сумму целых значений аргумента, для которых график функции + $y = \\dfrac{2x-1}{x^2+x-12}$ расположен выше прямой $y = \\dfrac{1}{2}$.`, + sol: `Условие: $\\dfrac{2x-1}{x^2+x-12} \\gt \\dfrac{1}{2}.$ Перенесём всё в одну часть: +$$\\dfrac{2x-1}{x^2+x-12} - \\dfrac{1}{2} \\gt 0 \\iff \\dfrac{2(2x-1)-(x^2+x-12)}{2(x^2+x-12)} \\gt 0 \\iff \\dfrac{-x^2+3x+10}{2(x^2+x-12)} \\gt 0.$$ +Умножим числитель и знаменатель на $-1$ (знак меняется): +$$\\dfrac{x^2-3x-10}{2(x^2+x-12)} \\lt 0.$$ +Разложим: $x^2-3x-10=(x-5)(x+2)$, $\\;x^2+x-12=(x+4)(x-3)$: +$$\\dfrac{(x-5)(x+2)}{2(x+4)(x-3)} \\lt 0.$$ +Критические точки: $-4,\\;-2,\\;3,\\;5$ (точки $-4$ и $3$ исключены из ОДЗ). +
Метод интервалов: + + + +
интервал$x\\lt-4$$(-4;-2)$$(-2;3)$$(3;5)$$x\\gt 5$
знак дроби$+$$-$$+$$-$$+$
+Решение: $x\\in(-4;\\;-2)\\cup(3;\\;5)$. +
Целые значения: в $(-4;-2)$ — это $-3$; в $(3;5)$ — это $4$. +
Сумма: $-3+4=1$. +
Ответ: $1$.
` + }, + { + text: `Дана окружность, длина которой равна $20\\pi$. + Найдите площадь сектора круга, ограниченного этой окружностью, + если угол этого сектора равен $72^{\\circ}$.`, + sol: `Формула длины окружности: $C = 2\\pi R$. +
Формула площади сектора с центральным углом $\\alpha^{\\circ}$: $S_{\\text{сект}} = \\dfrac{\\alpha}{360^{\\circ}}\\cdot \\pi R^{2}$. +
Шаг 1. Найдём радиус. По условию длина окружности равна $20\\pi$, значит: +$$2\\pi R = 20\\pi \\implies R = 10\\text{ см}.$$ +Шаг 2. Подставим в формулу площади сектора $\\alpha = 72^{\\circ}$ и $R = 10$: +$$S_{\\text{сект}} = \\dfrac{72}{360}\\cdot \\pi\\cdot 10^{2} = \\dfrac{1}{5}\\cdot 100\\pi = 20\\pi.$$ +
Ответ: $20\\pi$ (кв. ед.).
` + }, + { + text: `На соревнованиях авиамоделистов первая модель пролетела на $10\\%$, + или на $480$ м, меньше второй. Скорость первой модели на $20\\%$, + или на $1$ м/с, больше скорости второй модели. + Сколько минут находилась в воздухе каждая модель?`, + sol: `Связь процентов и десятичной дроби: $10\\%=0{,}1$, $20\\%=0{,}2$. +
Формула пути: $S=v\\cdot t$, откуда $t=\\dfrac{S}{v}$. +
Шаг 1. Найдём путь второй модели. По условию $10\\%$ от $S_{2}$ — это и есть $480$ м (разница между путями моделей). Составим уравнение: +$$0{,}1\\cdot S_{2} = 480 \\implies S_{2} = \\dfrac{480}{0{,}1} = 4800\\text{ м}.$$ +Первая модель пролетела на $480$ м меньше: +$$S_{1} = S_{2} - 480 = 4800 - 480 = 4320\\text{ м}.$$ +Шаг 2. Найдём скорость второй модели. Аналогично, $20\\%$ от $v_{2}$ равны $1$ м/с: +$$0{,}2\\cdot v_{2} = 1 \\implies v_{2} = \\dfrac{1}{0{,}2} = 5\\text{ м/с}.$$ +Скорость первой модели больше на $1$ м/с: +$$v_{1} = v_{2} + 1 = 6\\text{ м/с}.$$ +Шаг 3. Найдём время полёта каждой модели и переведём в минуты ($60$ с $= 1$ мин): +$$t_{1} = \\dfrac{S_{1}}{v_{1}} = \\dfrac{4320}{6} = 720\\text{ с} = 12\\text{ мин};$$ +$$t_{2} = \\dfrac{S_{2}}{v_{2}} = \\dfrac{4800}{5} = 960\\text{ с} = 16\\text{ мин}.$$ +
Ответ: 1-я модель — $12$ мин, 2-я модель — $16$ мин.
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v67.js b/frontend/js/exam9/variants/v67.js new file mode 100644 index 0000000..1a6a033 --- /dev/null +++ b/frontend/js/exam9/variants/v67.js @@ -0,0 +1,178 @@ +VARIANTS[67] = { + label: "Вариант 67", + tasks: [ + { + text: `Какое из данных чисел является простым:`, + opts: [ + ["а", "$9$"], ["б", "$1$"], ["в", "$77$"], ["г", "$51$"], ["д", "$2$"], + ], + sol: `

Проверяем каждое число:

+
    +
  • $9 = 3 \\times 3$ — составное;
    • +
  • $1$ — не является ни простым, ни составным по определению;
    • +
  • $77 = 7 \\times 11$ — составное;
    • +
  • $51 = 3 \\times 17$ — составное;
    • +
  • $2$ — делится только на $1$ и на себя, значит простое.
    • +
+
Ответ: д) $2$
` + }, + { + text: `Абсцисса точки, принадлежащей графику функции $y = -3x + 2$, равна $1$. + Тогда ордината этой точки равна:`, + opts: [ + ["а", "$5$"], ["б", "$-1$"], ["в", "$1$"], ["г", "$-1{,}5$"], ["д", "$\\dfrac{2}{3}$"], + ], + sol: `

Подставляем $x = 1$ в формулу функции:

+

$$y = -3 \\cdot 1 + 2 = -3 + 2 = -1.$$

+
Ответ: б) $-1$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "для прямоугольного треугольника $ABC$ с прямым углом $C$ верно $\\sin A = \\dfrac{BC}{AB}$;"], + ["б", "диагонали прямоугольника равны;"], + ["в", "площадь трапеции равна произведению полусуммы оснований на высоту;"], + ["г", "сумма градусных мер всех углов квадрата равна $180^{\\circ}$?"], + ], + sol: `

Квадрат — это четырёхугольник, сумма внутренних углов которого равна $360^{\\circ}$, а не $180^{\\circ}$.

+

Утверждения а), б), в) — верны. Утверждение г) — неверно.

+
Ответ: г)
` + }, + { + text: `Решите уравнение $\\dfrac{x}{-3{,}6} = \\dfrac{0{,}25}{-0{,}9}$.`, + sol: `

Из свойства пропорции $\\dfrac{x}{-3{,}6} = \\dfrac{0{,}25}{-0{,}9}$:

+

$$x = \\frac{(-3{,}6) \\cdot 0{,}25}{-0{,}9} = \\frac{-0{,}9}{-0{,}9} = 1.$$

+
Ответ: $x = 1$
` + }, + { + text: `Найдите сумму натуральных значений переменной из области определения + выражения $\\sqrt{-2x+6}$.`, + sol: `Условие существования квадратного корня: $\\sqrt{f(x)}$ определён только тогда, когда подкоренное выражение неотрицательно, то есть $f(x) \\geq 0$. +
Шаг 1. Запишем условие для нашего выражения $\\sqrt{-2x+6}$: +$$-2x + 6 \\geq 0.$$ +Шаг 2. Решим неравенство. Перенесём $-2x$ в правую часть: +$$6 \\geq 2x,$$ +а затем разделим обе части на $2$ (положительное число — знак не меняется): +$$x \\leq 3.$$ +Шаг 3. Выберем натуральные числа из найденной области. Натуральные числа — это $1,\\;2,\\;3,\\;4,\\ldots$ Из них условию $x \\leq 3$ удовлетворяют: $1,\\;2,\\;3$. +
Шаг 4. Найдём их сумму: +$$1 + 2 + 3 = 6.$$ +
Ответ: $6$
` + }, + { + text: `Найдите шестой член арифметической прогрессии, если её третий член равен $9$, + а разность прогрессии равна $-2$.`, + sol: `Формула $n$-го члена арифметической прогрессии: $a_n = a_1 + (n-1)d$. +
Из неё легко получить связь любых двух членов: $a_n = a_k + (n-k)d$, так как от $k$-го члена до $n$-го нужно прибавить разность $d$ ровно $(n-k)$ раз. +
Шаг 1. По условию $a_3 = 9$, $d = -2$. Найдём $a_6$, прибавив разность $d$ три раза (от 3-го к 6-му члену): +$$a_6 = a_3 + (6-3)\\cdot d = a_3 + 3d.$$ +Шаг 2. Подставим значения: +$$a_6 = 9 + 3\\cdot(-2) = 9 - 6 = 3.$$ +
Ответ: $3$
` + }, + { + text: `В параллелограмм с диагоналями, равными $6$ см и $8$ см, вписана окружность. + Найдите радиус этой окружности.`, + sol: `

Параллелограмм с вписанной окружностью является ромбом (суммы противоположных сторон равны, что в параллелограмме означает равенство всех сторон).

+

Полудиагонали ромба: $d_1/2 = 3$ см, $d_2/2 = 4$ см. Сторона ромба:

+

$$a = \\sqrt{3^2 + 4^2} = \\sqrt{9 + 16} = \\sqrt{25} = 5 \\text{ см}.$$

+

Площадь ромба:

+

$$S = \\frac{d_1 \\cdot d_2}{2} = \\frac{6 \\cdot 8}{2} = 24 \\text{ см}^2.$$

+

Полупериметр: $p = 2a = 2 \\cdot 5 = 10$ см. Радиус вписанной окружности:

+

$$r = \\frac{S}{p} = \\frac{24}{10} = 2{,}4 \\text{ см}.$$

+ + + + + + + + + + + + + + + A + B + C + D + + d₁=6 + d₁=6 + d₂=8 + d₂=8 + + a=5 + r=2,4 + +
Ответ: $r = 2{,}4$ см
` + }, + { + text: `Найдите значение выражения + $-\\dfrac{9}{5} : \\left(\\dfrac{16}{25} - 1\\right) - 0{,}025 : 0{,}01 + \\dfrac{1}{3} \\cdot (-6) - 4 : \\dfrac{2}{5}$. + В ответ запишите число, обратное ему.`, + sol: `Порядок действий: сначала выполняем действия в скобках, затем умножение и деление, в конце — сложение и вычитание (слева направо). +
Правило деления дробей: $\\dfrac{a}{b}:\\dfrac{c}{d}=\\dfrac{a}{b}\\cdot\\dfrac{d}{c}$. +
Обратное число к ненулевому числу $a$ — это число $\\dfrac{1}{a}$. У дроби $\\dfrac{p}{q}$ обратное равно $\\dfrac{q}{p}$. +
Шаг 1. Вычислим выражение в скобках, приведя $1$ к знаменателю $25$: +$$\\dfrac{16}{25} - 1 = \\dfrac{16-25}{25} = -\\dfrac{9}{25}.$$ +Шаг 2. Выполним первое деление, заменив деление умножением на обратную дробь: +$$-\\dfrac{9}{5} : \\left(-\\dfrac{9}{25}\\right) = -\\dfrac{9}{5}\\cdot\\left(-\\dfrac{25}{9}\\right) = \\dfrac{9\\cdot 25}{5\\cdot 9} = 5.$$ +Шаг 3. Выполним второе деление десятичных дробей: +$$0{,}025 : 0{,}01 = \\dfrac{0{,}025}{0{,}01} = 2{,}5.$$ +Шаг 4. Вычислим оставшиеся произведение и частное: +$$\\dfrac{1}{3}\\cdot(-6) = -2,\\qquad 4:\\dfrac{2}{5} = 4\\cdot\\dfrac{5}{2} = 10.$$ +Шаг 5. Соберём всё вместе: +$$5 - 2{,}5 + (-2) - 10 = 5 - 2{,}5 - 2 - 10 = -9{,}5 = -\\dfrac{19}{2}.$$ +Шаг 6. Запишем число, обратное полученному: у дроби $-\\dfrac{19}{2}$ обратная равна $-\\dfrac{2}{19}$. +
Ответ: $-\\dfrac{2}{19}$
` + }, + { + text: `Первую половину пути в $20$ км пешеход преодолел со скоростью на $10\\%$ меньше планируемой, + а вторую половину пути — со скоростью на $10\\%$ больше, чем планировал. + Как изменится время его движения по сравнению с планируемым?`, + sol: `

Пусть плановая скорость равна $v$. Половина пути — $10$ км.

+

Фактическое время:

+

$$t = \\frac{10}{0{,}9v} + \\frac{10}{1{,}1v} = \\frac{10}{v}\\left(\\frac{1}{0{,}9} + \\frac{1}{1{,}1}\\right) = \\frac{10}{v} \\cdot \\frac{1{,}1 + 0{,}9}{0{,}99} = \\frac{10}{v} \\cdot \\frac{2}{0{,}99} = \\frac{20}{0{,}99v}.$$

+

Плановое время: $t_0 = \\dfrac{20}{v}$.

+

Отношение: $\\dfrac{t}{t_0} = \\dfrac{20/(0{,}99v)}{20/v} = \\dfrac{1}{0{,}99} = \\dfrac{100}{99} \\gt 1$.

+

Фактическое время увеличится примерно на $\\dfrac{1}{99} \\approx 1\\%$ от планируемого.

+
Ответ: время увеличится (на $\\dfrac{1}{99}$ часть от планируемого, примерно на $1\\%$)
` + }, + { + text: `В треугольник $ABC$ вписан квадрат, две вершины которого лежат на стороне $AC$, + две другие — на сторонах $AB$ и $BC$. + Найдите площадь квадрата, если $AC = 30$ см, высота треугольника $BH = 20$ см.`, + sol: `

Пусть сторона квадрата равна $a$. Квадрат расположен основанием на $AC$.

+

На высоте $a$ от $AC$ ширина треугольника (по подобию) равна:

+

$$AC \\cdot \\frac{BH - a}{BH} = 30 \\cdot \\frac{20 - a}{20}.$$

+

Эта ширина должна равняться стороне квадрата $a$:

+

$$30 \\cdot \\frac{20 - a}{20} = a \\Rightarrow \\frac{3(20 - a)}{2} = a \\Rightarrow 60 - 3a = 2a \\Rightarrow 5a = 60 \\Rightarrow a = 12.$$

+

Площадь квадрата: $S = a^2 = 12^2 = 144$ см².

+ + + + + + + + + + + + + A + B + C + H + + a=12 + AC = 30 см + BH=20 + +
Ответ: $144$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v68.js b/frontend/js/exam9/variants/v68.js new file mode 100644 index 0000000..b66992a --- /dev/null +++ b/frontend/js/exam9/variants/v68.js @@ -0,0 +1,164 @@ +VARIANTS[68] = { + label: "Вариант 68", + tasks: [ + { + text: `Какое из данных чисел является простым:`, + opts: [ + ["а", "$39$"], ["б", "$6$"], ["в", "$7$"], ["г", "$1$"], ["д", "$15$"], + ], + sol: `

Проверяем каждое число:

+
    +
  • $39 = 3 \\times 13$ — составное;
    • +
  • $6 = 2 \\times 3$ — составное;
    • +
  • $7$ — делится только на $1$ и на себя, значит простое;
    • +
  • $1$ — не является ни простым, ни составным по определению;
    • +
  • $15 = 3 \\times 5$ — составное.
    • +
+
Ответ: в) $7$
` + }, + { + text: `Абсцисса точки, принадлежащей графику функции $y = 3x - 2$, равна $-1$. + Тогда ордината этой точки равна:`, + opts: [ + ["а", "$5$"], ["б", "$-5$"], ["в", "$-1$"], ["г", "$\\dfrac{2}{3}$"], ["д", "$\\dfrac{1}{3}$"], + ], + sol: `

Подставляем $x = -1$ в формулу функции:

+

$$y = 3 \\cdot (-1) - 2 = -3 - 2 = -5.$$

+
Ответ: б) $-5$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "для прямоугольного треугольника $ABC$ с прямым углом $C$ верно $\\cos A = \\dfrac{AC}{AB}$;"], + ["б", "диагонали ромба взаимно перпендикулярны;"], + ["в", "площадь ромба равна половине произведения диагоналей;"], + ["г", "сумма градусных мер всех углов квадрата равна $270^{\\circ}$?"], + ], + sol: `

Квадрат — четырёхугольник, сумма внутренних углов которого равна $360^{\\circ}$, а не $270^{\\circ}$.

+

Утверждения а), б), в) — верны. Утверждение г) — неверно.

+
Ответ: г)
` + }, + { + text: `Решите уравнение $\\dfrac{x}{-5{,}6} = \\dfrac{-0{,}25}{-7}$.`, + sol: `

Из свойства пропорции $\\dfrac{x}{-5{,}6} = \\dfrac{-0{,}25}{-7} = \\dfrac{0{,}25}{7}$:

+

$$x = (-5{,}6) \\cdot \\dfrac{0{,}25}{7} = \\dfrac{-5{,}6 \\cdot 0{,}25}{7} = \\dfrac{-1{,}4}{7} = -0{,}2.$$

+
Ответ: $x = -0{,}2$
` + }, + { + text: `Найдите сумму натуральных значений переменной из области определения + выражения $\\sqrt{9-3x}$.`, + sol: `Условие существования квадратного корня: $\\sqrt{f(x)}$ определён только тогда, когда подкоренное выражение неотрицательно, то есть $f(x) \\geq 0$. +
Шаг 1. Запишем условие для выражения $\\sqrt{9-3x}$: +$$9 - 3x \\geq 0.$$ +Шаг 2. Перенесём $-3x$ в правую часть: +$$9 \\geq 3x.$$ +Шаг 3. Разделим обе части на $3$ (положительное число — знак не меняется): +$$3 \\geq x, \\quad \\text{то есть} \\quad x \\leq 3.$$ +Шаг 4. Натуральные числа из области определения — те, что не превосходят $3$: это $1,\\;2,\\;3$. +
Шаг 5. Их сумма: $1 + 2 + 3 = 6$. +
Ответ: $6$
` + }, + { + text: `Найдите шестой член арифметической прогрессии, если её второй член равен $5$, + а разность прогрессии равна $2$.`, + sol: `Формула $n$-го члена арифметической прогрессии: $a_n = a_1 + (n-1)d$. +
Связь любых двух членов: $a_n = a_k + (n-k)d$ — от $k$-го члена до $n$-го прибавляем разность $(n-k)$ раз. +
Шаг 1. По условию $a_2 = 5$, $d = 2$. От второго члена до шестого нужно прибавить разность $6 - 2 = 4$ раза: +$$a_6 = a_2 + (6 - 2)\\cdot d = a_2 + 4d.$$ +Шаг 2. Подставим значения: +$$a_6 = 5 + 4\\cdot 2 = 5 + 8 = 13.$$ +
Ответ: $13$
` + }, + { + text: `В параллелограмм с диагоналями, равными $10$ см и $24$ см, вписана окружность. + Найдите радиус этой окружности.`, + sol: `Свойство описанного четырёхугольника: если в четырёхугольник вписана окружность, то суммы его противоположных сторон равны. +
В параллелограмме противоположные стороны и так равны, поэтому условие $AB+CD = BC+AD$ даёт $2AB = 2BC$, то есть все стороны равны. Значит, такой параллелограмм — это ромб. +
Свойство диагоналей ромба: диагонали ромба взаимно перпендикулярны и точкой пересечения делятся пополам. +
Шаг 1. Найдём сторону ромба. Половины диагоналей $\\dfrac{d_1}{2} = 5$ см и $\\dfrac{d_2}{2} = 12$ см являются катетами прямоугольного треугольника, гипотенуза которого — сторона ромба. По теореме Пифагора: +$$a = \\sqrt{5^2 + 12^2} = \\sqrt{25 + 144} = \\sqrt{169} = 13\\text{ см}.$$ +Шаг 2. Найдём площадь ромба. Формула площади ромба через диагонали: $S = \\dfrac{d_1 \\cdot d_2}{2}$: +$$S = \\dfrac{10 \\cdot 24}{2} = 120\\text{ см}^{2}.$$ +Шаг 3. Найдём радиус вписанной окружности. Формула: $r = \\dfrac{S}{p}$, где $p$ — полупериметр. +
Так как у ромба все четыре стороны равны $a$, полупериметр: $p = \\dfrac{4a}{2} = 2a = 26$ см. +$$r = \\dfrac{S}{p} = \\dfrac{120}{26} = \\dfrac{60}{13}\\text{ см} \\approx 4{,}6\\text{ см}.$$ + + + + + + + + + A + B + C + D + d₁=24 + d₁=24 + d₂=10 + d₂=10 + a=13 + r=60/13 + +
Ответ: $r = \\dfrac{60}{13}$ см
` + }, + { + text: `Найдите значение выражения + $-8 : \\left(-1\\dfrac{1}{7}\\right) + \\dfrac{1}{2} \\cdot \\left(-1\\dfrac{1}{5}\\right) + 6 \\cdot \\left(-\\dfrac{1}{3}\\right) - 8 : \\dfrac{4}{5}$. + В ответ запишите число, обратное ему.`, + sol: `Перевод смешанного числа в неправильную дробь: $a\\dfrac{b}{c}=\\dfrac{a\\cdot c+b}{c}$. +
Правило деления дробей: $\\dfrac{a}{b}:\\dfrac{c}{d}=\\dfrac{a}{b}\\cdot\\dfrac{d}{c}$. +
Обратное число к дроби $\\dfrac{p}{q}$ равно $\\dfrac{q}{p}$. +
Шаг 1. Переведём смешанные числа в неправильные дроби: +$$1\\dfrac{1}{7} = \\dfrac{8}{7},\\qquad 1\\dfrac{1}{5} = \\dfrac{6}{5}.$$ +Шаг 2. Вычислим первое деление, заменив его умножением на обратную дробь: +$$-8 : \\left(-\\dfrac{8}{7}\\right) = -8\\cdot\\left(-\\dfrac{7}{8}\\right) = 7.$$ +Шаг 3. Вычислим первое произведение: +$$\\dfrac{1}{2}\\cdot\\left(-\\dfrac{6}{5}\\right) = -\\dfrac{6}{10} = -\\dfrac{3}{5}.$$ +Шаг 4. Вычислим оставшиеся произведение и частное: +$$6\\cdot\\left(-\\dfrac{1}{3}\\right) = -2,\\qquad 8 : \\dfrac{4}{5} = 8\\cdot\\dfrac{5}{4} = 10.$$ +Шаг 5. Соберём всё вместе и приведём к общему знаменателю $5$: +$$7 - \\dfrac{3}{5} - 2 - 10 = -5 - \\dfrac{3}{5} = -\\dfrac{25}{5} - \\dfrac{3}{5} = -\\dfrac{28}{5}.$$ +Шаг 6. Запишем обратное число: для $-\\dfrac{28}{5}$ обратное равно $-\\dfrac{5}{28}$. +
Ответ: $-\\dfrac{5}{28}$
` + }, + { + text: `Первую половину пути в $120$ км велосипедист преодолел со скоростью на $20\\%$ меньше планируемой, + а вторую половину пути — со скоростью на $20\\%$ больше, чем планировал. + Как изменится время его движения по сравнению с планируемым?`, + sol: `

Пусть плановая скорость равна $v$. Половина пути — $60$ км.

+

Фактическое время:

+

$$t = \\dfrac{60}{0{,}8v} + \\dfrac{60}{1{,}2v} = \\dfrac{60}{v}\\left(\\dfrac{1}{0{,}8} + \\dfrac{1}{1{,}2}\\right) = \\dfrac{60}{v} \\cdot \\dfrac{1{,}2 + 0{,}8}{0{,}96} = \\dfrac{60}{v} \\cdot \\dfrac{2}{0{,}96} = \\dfrac{120}{0{,}96v} = \\dfrac{125}{v}.$$

+

Плановое время: $t_0 = \\dfrac{120}{v}$.

+

Отношение: $\\dfrac{t}{t_0} = \\dfrac{125/v}{120/v} = \\dfrac{125}{120} = \\dfrac{25}{24} \\gt 1$.

+

Фактическое время увеличится на $\\dfrac{25}{24} - 1 = \\dfrac{1}{24}$ часть от планируемого (примерно на $4{,}2\\%$).

+
Ответ: время увеличится (на $\\dfrac{1}{24}$ часть от планируемого, примерно на $4{,}2\\%$)
` + }, + { + text: `В треугольник $ABC$ вписан квадрат, две вершины которого лежат на стороне $AC$, + две другие — на сторонах $AB$ и $BC$. + Найдите площадь квадрата, если $AC = 60$ см, высота треугольника $BH = 30$ см.`, + sol: `

Пусть сторона квадрата равна $a$. Квадрат расположен основанием на $AC$.

+

На высоте $a$ от $AC$ ширина треугольника (по подобию) равна:

+

$$AC \\cdot \\dfrac{BH - a}{BH} = 60 \\cdot \\dfrac{30 - a}{30} = 2(30 - a).$$

+

Эта ширина должна равняться стороне квадрата $a$:

+

$$2(30 - a) = a \\Rightarrow 60 - 2a = a \\Rightarrow 3a = 60 \\Rightarrow a = 20.$$

+

Площадь квадрата: $S = a^2 = 20^2 = 400$ см².

+ + + + + + A + B + C + H + a=20 + AC = 60 см + BH=30 + +
Ответ: $400$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v69.js b/frontend/js/exam9/variants/v69.js new file mode 100644 index 0000000..8e6b775 --- /dev/null +++ b/frontend/js/exam9/variants/v69.js @@ -0,0 +1,163 @@ +VARIANTS[69] = { + label: "Вариант 69", + tasks: [ + { + text: `Определите, какое из данных равенств является верным:`, + opts: [ + ["а", "$3(x-y) = 3x-y$"], ["б", "$3(x-y) = x-3y$"], ["в", "$3(x-y) = 3x-3y$"], + ["г", "$3(x-y) = 3y-3x$"], ["д", "$3(x-y) = 3x+3y$"], + ], + sol: `По распределительному закону умножения: +$$3(x-y) = 3\\cdot x - 3\\cdot y = 3x-3y$$ +
    +
  • а) $3x-y$ — вынесли тройку только из первого слагаемого — неверно
    • +
  • б) $x-3y$ — вынесли тройку только из второго слагаемого — неверно
    • +
  • в) $3x-3y$ — верно ✓
    • +
  • г) $3y-3x$ — знак изменён — неверно
    • +
  • д) $3x+3y$ — знак минус заменён на плюс — неверно
    • +
+
Ответ: в)
` + }, + { + text: `$30\\%$ от числа $120$ равны:`, + opts: [ + ["а", "$3{,}6$"], ["б", "$360$"], ["в", "$36$"], ["г", "$150$"], ["д", "$400$"], + ], + sol: `$$120 \\cdot 0{,}3 = 36$$ +
Ответ: в) $36$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "медианы треугольника пересекаются в одной точке;"], + ["б", "радиус вписанной в треугольник окружности можно найти по формуле $r = S/p$;"], + ["в", "в треугольнике против большей стороны лежит больший угол;"], + ["г", "в любой параллелограмм можно вписать окружность?"], + ], + sol: `
    +
  • а) Медианы треугольника пересекаются в одной точке — верно
    • +
  • б) $r = S/p$ (где $p$ — полупериметр) — верно
    • +
  • в) Против большей стороны лежит больший угол — верно
    • +
  • г) В любой параллелограмм можно вписать окружность — НЕВЕРНО
    • +
+Для вписанной окружности нужно: $AB+CD = BC+AD$. В параллелограмме $AB=CD$ и $BC=AD$, поэтому условие даёт $AB=BC$ — это ромб. Обычный параллелограмм (не ромб) вписанной окружности не имеет. +
Ответ: г)
` + }, + { + text: `Найдите все целые решения неравенства $-23 \\leq 10x \\leq 13$.`, + sol: `Делим все части на $10$: +$$-2{,}3 \\leq x \\leq 1{,}3$$ +Целые числа на отрезке $[-2{,}3;\\; 1{,}3]$: это $-2,\\; -1,\\; 0,\\; 1$. +
Ответ: $-2,\\; -1,\\; 0,\\; 1$
` + }, + { + text: `Найдите четвёртый член геометрической прогрессии, если её первый член равен $5$, + а знаменатель прогрессии равен $2$.`, + sol: `Формула $n$-го члена геометрической прогрессии: $a_n = a_1 \\cdot q^{n-1}$, где $a_1$ — первый член, $q$ — знаменатель прогрессии. +
Шаг 1. По условию $a_1 = 5$, $q = 2$, нужно найти $a_4$. Подставляем $n = 4$: +$$a_4 = a_1 \\cdot q^{4-1} = 5 \\cdot 2^{3}.$$ +Шаг 2. Вычислим $2^3 = 8$ и подставим: +$$a_4 = 5 \\cdot 8 = 40.$$ +
Ответ: $40$
` + }, + { + text: `Найдите синус угла $BAC$, изображённого на клетчатой бумаге.`, + figure: ``, + sol: `Определение синуса в прямоугольном треугольнике: $\\sin\\alpha = \\dfrac{\\text{противолежащий катет}}{\\text{гипотенуза}}$. +
Теорема Пифагора: для прямоугольного треугольника с катетами $a$, $b$ и гипотенузой $c$ верно $c^2 = a^2 + b^2$. +
Шаг 1. По рисунку достроим прямоугольный треугольник так, чтобы $\\angle BAC$ стал острым углом этого треугольника, а катеты шли по линиям клеток. +
Шаг 2. Посчитаем длины катетов по клеткам. +
Шаг 3. По теореме Пифагора находим гипотенузу. +
Шаг 4. Применяем формулу синуса: делим длину противолежащего катета на длину гипотенузы. +
Ответ: определяется по рисунку
` + }, + { + text: `Решите уравнение $(x-4)^2 - (x+6)^2 = 30$.`, + sol: `Формулы сокращённого умножения (квадрат разности и квадрат суммы): +
$(a-b)^2 = a^2 - 2ab + b^2$,  $(a+b)^2 = a^2 + 2ab + b^2$. +
Шаг 1. Раскроем квадраты: +$$(x-4)^2 = x^2 - 8x + 16,$$ +$$(x+6)^2 = x^2 + 12x + 36.$$ +Шаг 2. Подставим в уравнение и аккуратно раскроем скобки (минус перед скобкой меняет знаки всех слагаемых): +$$(x^2 - 8x + 16) - (x^2 + 12x + 36) = 30,$$ +$$x^2 - 8x + 16 - x^2 - 12x - 36 = 30.$$ +Шаг 3. Приведём подобные слагаемые ($x^2 - x^2 = 0$, $-8x - 12x = -20x$, $16 - 36 = -20$): +$$-20x - 20 = 30.$$ +Шаг 4. Перенесём $-20$ в правую часть, поменяв знак: +$$-20x = 50.$$ +Шаг 5. Разделим обе части на $-20$ (знак меняется только когда делим неравенство, в уравнении знак сохраняется): +$$x = \\dfrac{50}{-20} = -2{,}5.$$ +
Ответ: $x = -2{,}5$
` + }, + { + text: `Упростите выражение + $\\dfrac{(2x^2+1)(4x^4+1)(16x^8+1)(2x^2-1)}{256x^{16}-1}$.`, + sol: `Формула разности квадратов: $a^2-b^2=(a-b)(a+b)$, или, наоборот, $(a-b)(a+b)=a^2-b^2$. +
Шаг 1. Воспользуемся переместительным свойством умножения и поставим $(2x^2-1)$ рядом с $(2x^2+1)$. Применим формулу разности квадратов: +$$(2x^2-1)(2x^2+1) = (2x^2)^2 - 1^2 = 4x^4 - 1.$$ +Шаг 2. Теперь рядом с $(4x^4+1)$ стоит множитель $(4x^4-1)$. Снова применим формулу разности квадратов: +$$(4x^4-1)(4x^4+1) = (4x^4)^2 - 1 = 16x^8 - 1.$$ +Шаг 3. Аналогично умножим на $(16x^8+1)$: +$$(16x^8-1)(16x^8+1) = (16x^8)^2 - 1 = 256x^{16} - 1.$$ +Шаг 4. Получили, что числитель равен $256x^{16}-1$ — это в точности совпадает со знаменателем. Значит, дробь равна единице: +$$\\dfrac{256x^{16}-1}{256x^{16}-1} = 1.$$ +
Ответ: $1$
` + }, + { + text: `Брат и сестра вышли одновременно из дома в тренажёрный зал, находящийся на расстоянии + $1$ км $250$ м от дома. Дойдя до тренажёрного зала, брат вспомнил, что забыл абонемент, + и с той же скоростью отправился домой. На каком расстоянии от тренажёрного зала + брат встретит сестру, если скорость брата $5$ км/ч, а скорость сестры $3$ км/ч?`, + sol: `Расстояние $d = 1{,}25$ км. Скорость брата $5$ км/ч, сестры $3$ км/ч. +
Шаг 1. Время брата до зала: $t_1 = \\dfrac{1{,}25}{5} = 0{,}25$ ч. +
Шаг 2. За это время сестра прошла $3 \\cdot 0{,}25 = 0{,}75$ км. До зала осталось: $1{,}25 - 0{,}75 = 0{,}5$ км. +
Шаг 3. Брат выходит из зала навстречу. Скорость сближения $5 + 3 = 8$ км/ч. Время: $t_2 = \\dfrac{0{,}5}{8} = \\dfrac{1}{16}$ ч. +
Шаг 4. Брат прошёл от зала: $5 \\cdot \\dfrac{1}{16} = \\dfrac{5}{16}$ км $= 312{,}5$ м. +
Ответ: $312{,}5$ м от тренажёрного зала
` + }, + { + text: `В треугольнике $ABC$ медиана $AM$ перпендикулярна биссектрисе $BK$. + Найдите длину стороны $AB$, если $AM = BK = 20$.`, + sol: ` + + + + + + + + + + + + + + + + + A + B + C + M + K + P + + AM=20 + BK=20 + + 10 + 10 + 15 + 5 + +Шаг 1. Ключевые отрезки. +
Пусть $P$ — точка пересечения $AM$ и $BK$. Поскольку $AM\\perp BK$ и $AM=BK=20$, из свойств медианы и биссектрисы в таком треугольнике можно показать, что: +$$AP = PM = 10\\text{ (медиана делится пополам)}$$ +$$BP = 15,\\quad PK = 5\\text{ (биссектриса делится в отношении 3:1)}$$ +Шаг 2. Теорема Пифагора в $\\triangle APB$. +
$\\angle APB = 90°$ (медиана $\\perp$ биссектрисе), катеты $AP=10$ и $BP=15$: +$$AB = \\sqrt{AP^2 + BP^2} = \\sqrt{10^2 + 15^2} = \\sqrt{100 + 225} = \\sqrt{325} = 5\\sqrt{13}$$ +
Ответ: $AB = 5\\sqrt{13}$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v70.js b/frontend/js/exam9/variants/v70.js new file mode 100644 index 0000000..3265ed0 --- /dev/null +++ b/frontend/js/exam9/variants/v70.js @@ -0,0 +1,167 @@ +VARIANTS[70] = { + label: "Вариант 70", + tasks: [ + { + text: `Определите, какое из данных равенств является верным:`, + opts: [ + ["а", "$4(x-y) = x-4y$"], ["б", "$4(x-y) = 4x-y$"], ["в", "$4(x-y) = 4x-4y$"], + ["г", "$4(x-y) = 4y-4x$"], ["д", "$4(x-y) = 4x+4y$"], + ], + sol: `По распределительному закону умножения: +$$4(x-y) = 4\\cdot x - 4\\cdot y = 4x-4y$$ +
    +
  • а) $x-4y$ — четвёрка вынесена только из второго слагаемого — неверно
    • +
  • б) $4x-y$ — четвёрка вынесена только из первого слагаемого — неверно
    • +
  • в) $4x-4y$ — верно ✓
    • +
  • г) $4y-4x$ — знак изменён — неверно
    • +
  • д) $4x+4y$ — знак минус заменён на плюс — неверно
    • +
+
Ответ: в)
` + }, + { + text: `$40\\%$ от числа $220$ равны:`, + opts: [ + ["а", "$8{,}8$"], ["б", "$260$"], ["в", "$88$"], ["г", "$80$"], ["д", "$550$"], + ], + sol: `$$220 \\cdot 0{,}4 = 88$$ +
Ответ: в) $88$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "биссектрисы треугольника пересекаются в одной точке;"], + ["б", "радиус окружности, описанной около треугольника, можно найти по формуле $R = \\dfrac{abc}{4S}$;"], + ["в", "в треугольнике против большего угла лежит большая сторона;"], + ["г", "около любого параллелограмма можно описать окружность?"], + ], + sol: `
    +
  • а) Биссектрисы треугольника пересекаются в одной точке — верно
    • +
  • б) $R = \\dfrac{abc}{4S}$ — стандартная формула радиуса описанной окружности — верно
    • +
  • в) Против большего угла лежит большая сторона — верно
    • +
  • г) Около любого параллелограмма можно описать окружность — НЕВЕРНО
    • +
+Для описанной окружности необходимо, чтобы сумма противоположных углов равнялась $180°$. В произвольном параллелограмме $\\angle A = \\angle C$ и $\\angle B = \\angle D$, поэтому $\\angle A + \\angle C = 2\\angle A \\neq 180°$ в общем случае. Описанная окружность существует лишь у прямоугольника. +
Ответ: г)
` + }, + { + text: `Найдите все целые решения неравенства $-9 \\leq 2x \\leq -3$.`, + sol: `Делим все части на $2$: +$$-4{,}5 \\leq x \\leq -1{,}5$$ +Целые числа на отрезке $[-4{,}5;\\; -1{,}5]$: это $-4,\\; -3,\\; -2$. +
Ответ: $-4,\\; -3,\\; -2$
` + }, + { + text: `Найдите третий член геометрической прогрессии, если её первый член равен $0{,}2$, + а знаменатель прогрессии равен $2{,}5$.`, + sol: `Формула $n$-го члена геометрической прогрессии: $a_n = a_1 \\cdot q^{n-1}$, где $a_1$ — первый член, $q$ — знаменатель прогрессии. +
Шаг 1. По условию $a_1 = 0{,}2$, $q = 2{,}5$, нужно найти $a_3$. Подставляем $n = 3$: +$$a_3 = a_1 \\cdot q^{3-1} = 0{,}2 \\cdot (2{,}5)^{2}.$$ +Шаг 2. Вычислим $(2{,}5)^2 = 6{,}25$: +$$a_3 = 0{,}2 \\cdot 6{,}25 = 1{,}25.$$ +
Ответ: $1{,}25$
` + }, + { + text: `Найдите косинус угла $ACB$, изображённого на клетчатой бумаге.`, + figure: ``, + sol: `Определение косинуса в прямоугольном треугольнике: $\\cos\\alpha = \\dfrac{\\text{прилежащий катет}}{\\text{гипотенуза}}$. +
Теорема Пифагора: для прямоугольного треугольника с катетами $a$, $b$ и гипотенузой $c$ верно $c^2 = a^2 + b^2$. +
Шаг 1. По рисунку достроим прямоугольный треугольник так, чтобы угол $\\angle ACB$ оказался острым углом этого треугольника, а катеты шли по линиям клеток. +
Шаг 2. Посчитаем длины катетов по клеткам. +
Шаг 3. По теореме Пифагора находим гипотенузу. +
Шаг 4. Применяем формулу косинуса: делим длину прилежащего катета на длину гипотенузы. +
Ответ: определяется по рисунку
` + }, + { + text: `Решите уравнение $(x-2)^2 - (x+8)^2 = 30$.`, + sol: `Формулы сокращённого умножения (квадрат разности и квадрат суммы): +
$(a-b)^2 = a^2 - 2ab + b^2$,  $(a+b)^2 = a^2 + 2ab + b^2$. +
Шаг 1. Раскроем квадраты: +$$(x-2)^2 = x^2 - 4x + 4,$$ +$$(x+8)^2 = x^2 + 16x + 64.$$ +Шаг 2. Подставим в уравнение, аккуратно раскроем скобки (минус перед скобкой меняет знаки): +$$(x^2 - 4x + 4) - (x^2 + 16x + 64) = 30,$$ +$$x^2 - 4x + 4 - x^2 - 16x - 64 = 30.$$ +Шаг 3. Приведём подобные слагаемые ($x^2 - x^2 = 0$, $-4x - 16x = -20x$, $4 - 64 = -60$): +$$-20x - 60 = 30.$$ +Шаг 4. Перенесём $-60$ в правую часть, поменяв знак: +$$-20x = 90.$$ +Шаг 5. Разделим обе части на $-20$: +$$x = \\dfrac{90}{-20} = -4{,}5.$$ +
Ответ: $x = -4{,}5$
` + }, + { + text: `Упростите выражение + $\\dfrac{(x^2+2)(x^4+4)(x^8+16)(x^2-2)}{x^{16}-256}$.`, + sol: `Формула разности квадратов: $(a-b)(a+b) = a^2 - b^2$. +
Шаг 1. Переставим множители в числителе так, чтобы рядом оказались $(x^2-2)$ и $(x^2+2)$. Применим формулу разности квадратов: +$$(x^2-2)(x^2+2) = (x^2)^2 - 2^2 = x^4 - 4.$$ +Шаг 2. Полученный множитель $(x^4-4)$ умножим на $(x^4+4)$ — снова разность квадратов: +$$(x^4-4)(x^4+4) = (x^4)^2 - 4^2 = x^8 - 16.$$ +Шаг 3. Аналогично: +$$(x^8-16)(x^8+16) = (x^8)^2 - 16^2 = x^{16} - 256.$$ +Шаг 4. Числитель равен $x^{16}-256$ — совпадает со знаменателем, значит дробь равна $1$: +$$\\dfrac{x^{16}-256}{x^{16}-256} = 1.$$ +
Ответ: $1$
` + }, + { + text: `Брат и сестра вышли одновременно из дома в тренажёрный зал, находящийся на расстоянии + $1$ км $200$ м от дома. Дойдя до тренажёрного зала, сестра вспомнила, что забыла абонемент, + и с той же скоростью отправилась домой. На каком расстоянии от тренажёрного зала + сестра встретит брата, если скорость брата $3$ км/ч, а скорость сестры $2{,}4$ км/ч?`, + sol: `Формула пути: $S = v\\cdot t$, откуда $t = \\dfrac{S}{v}$. +
Скорость сближения при движении навстречу равна сумме скоростей. +
Шаг 1. Переведём расстояние в единые единицы: $d = 1$ км $200$ м $= 1{,}2$ км. Обозначим скорость брата $v_{1} = 3$ км/ч, скорость сестры $v_{2} = 2{,}4$ км/ч (брат идёт быстрее, поэтому первым придёт в зал именно он, и затем повернёт назад). +
Шаг 2. Время до того, как первый дошёл до зала: +$$t_{1} = \\dfrac{d}{v_{1}} = \\dfrac{1{,}2}{3} = 0{,}4\\text{ ч}.$$ +Шаг 3. За это время сестра прошла $v_{2}\\cdot t_{1} = 2{,}4\\cdot 0{,}4 = 0{,}96$ км. Значит, до зала ей осталось: +$$1{,}2 - 0{,}96 = 0{,}24\\text{ км}.$$ +Шаг 4. Теперь они движутся навстречу друг другу. Скорость сближения: +$$v_{сбл} = v_{1} + v_{2} = 3 + 2{,}4 = 5{,}4\\text{ км/ч}.$$ +Время до встречи: +$$t_{2} = \\dfrac{0{,}24}{5{,}4} = \\dfrac{24}{540} = \\dfrac{2}{45}\\text{ ч}.$$ +Шаг 5. Расстояние от зала до места встречи равно пути, который прошёл вышедший из зала: +$$x = v_{1}\\cdot t_{2} = 3\\cdot\\dfrac{2}{45} = \\dfrac{6}{45} = \\dfrac{2}{15}\\text{ км} \\approx 133{,}3\\text{ м}.$$ +
Ответ: $\\dfrac{2}{15}$ км $\\approx 133$ м от тренажёрного зала
` + }, + { + text: `В треугольнике $ABC$ медиана $AK$ перпендикулярна биссектрисе $BM$. + Найдите длину стороны $AB$, если $AK = BM = 12$.`, + sol: ` + + + + + + + + + + + A + B + C + K + M + P + AK=12 + BM=12 + 6 + 6 + 9 + 3 + +Теорема Пифагора: в прямоугольном треугольнике с катетами $a$, $b$ и гипотенузой $c$ выполняется $c^2 = a^2 + b^2$. +
Свойство медианы: медиана $AK$ из вершины $A$ делит сторону $BC$ пополам ($BK = KC$). +
Свойство биссектрисы: биссектриса $BM$ из вершины $B$ делит сторону $AC$ в отношении $AM:MC = AB:BC$. +
Шаг 1. Пусть $P$ — точка пересечения медианы $AK$ и биссектрисы $BM$. По условию $AK \\perp BM$ и $AK = BM = 12$. +
В этой стандартной конфигурации (медиана из $A$ перпендикулярна биссектрисе из $B$) выполняются соотношения: +$$AP = PK = \\dfrac{AK}{2} = 6$$ (точка $P$ — середина медианы $AK$); +$$BP:PM = 3:1, \\quad \\text{то есть}\\quad BP = \\dfrac{3}{4}\\cdot 12 = 9,\\;\\; PM = 3.$$ +Шаг 2. Рассмотрим прямоугольный треугольник $APB$. Так как $AK \\perp BM$, то $\\angle APB = 90^\\circ$, а катеты — это $AP = 6$ и $BP = 9$. По теореме Пифагора: +$$AB^2 = AP^2 + BP^2 = 6^2 + 9^2 = 36 + 81 = 117.$$ +Шаг 3. Извлечём корень. Так как $117 = 9 \\cdot 13$: +$$AB = \\sqrt{117} = \\sqrt{9}\\cdot\\sqrt{13} = 3\\sqrt{13}.$$ +
Ответ: $AB = 3\\sqrt{13}$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v71.js b/frontend/js/exam9/variants/v71.js new file mode 100644 index 0000000..333fbcf --- /dev/null +++ b/frontend/js/exam9/variants/v71.js @@ -0,0 +1,177 @@ +VARIANTS[71] = { + label: "Вариант 71", + tasks: [ + { + text: `Какое из данных чисел является решением неравенства $2x \\geq -1$:`, + opts: [ + ["а", "$-3$"], ["б", "$-2$"], ["в", "$-1$"], ["г", "$-1{,}5$"], ["д", "$-0{,}5$"], + ], + sol: `Решаем неравенство: $2x \\geq -1 \\Rightarrow x \\geq -0{,}5$. +
    +
  • а) $-3 \\lt -0{,}5$ — не является решением;
    • +
  • б) $-2 \\lt -0{,}5$ — не является решением;
    • +
  • в) $-1 \\lt -0{,}5$ — не является решением;
    • +
  • г) $-1{,}5 \\lt -0{,}5$ — не является решением;
    • +
  • д) $-0{,}5 \\geq -0{,}5$ — является решением ✓
    • +
+
Ответ: д) $-0{,}5$
` + }, + { + text: `Какое из следующих выражений равно $a^7$:`, + opts: [ + ["а", "$(a^5)^2$"], ["б", "$(a^3)^4$"], ["в", "$a^3 \\cdot a^4$"], + ["г", "$a^{14}/a^2$"], ["д", "$a^{21}/a^3$"], + ], + sol: `При умножении степеней с одним основанием показатели складываются: +
    +
  • а) $(a^5)^2 = a^{10}$ — не $a^7$;
    • +
  • б) $(a^3)^4 = a^{12}$ — не $a^7$;
    • +
  • в) $a^3 \\cdot a^4 = a^{3+4} = a^7$ — верно ✓
    • +
  • г) $a^{14}/a^2 = a^{12}$ — не $a^7$;
    • +
  • д) $a^{21}/a^3 = a^{18}$ — не $a^7$.
    • +
+
Ответ: в) $a^3 \\cdot a^4$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "медиана треугольника соединяет вершину с серединой противолежащей стороны;"], + ["б", "у любого параллелограмма все углы равны;"], + ["в", "периметр ромба со стороной $a$ равен $P = 4a$;"], + ["г", "около любого треугольника можно описать окружность?"], + ], + sol: `
    +
  • а) Медиана соединяет вершину с серединой противолежащей стороны — верно;
    • +
  • б) У любого параллелограмма все углы равны — НЕВЕРНО. Все четыре угла равны лишь у прямоугольника. В общем параллелограмме два острых и два тупых угла;
    • +
  • в) Периметр ромба $P = 4a$ — верно;
    • +
  • г) Около любого треугольника можно описать окружность — верно.
    • +
+
Ответ: б)
` + }, + { + text: `Раскройте скобки и приведите подобные слагаемые + $2a^2(1-a) - (-2a^3 + 3a^2)$.`, + sol: `Раскрываем скобки: +$$2a^2(1-a) - (-2a^3+3a^2) = 2a^2 - 2a^3 + 2a^3 - 3a^2$$ +Приводим подобные слагаемые: +$$= (2a^2 - 3a^2) + (-2a^3 + 2a^3) = -a^2 + 0 = -a^2$$ +
Ответ: $-a^2$
` + }, + { + text: `Ордината точки, принадлежащей графику функции $y = 2x + 2$, + равна числу, противоположному числу $4$. + Найдите абсциссу этой точки.`, + sol: `Противоположное число: для числа $a$ противоположным называется число $-a$ (сумма $a+(-a)=0$). +
Связь координат точки с уравнением функции: точка $(x;y)$ принадлежит графику функции $y=f(x)$ тогда и только тогда, когда её координаты удовлетворяют уравнению. +
Шаг 1. По условию ордината точки — число, противоположное числу $4$. Значит, $y = -4$. +
Шаг 2. Подставим $y = -4$ в уравнение функции $y = 2x + 2$: +$$2x + 2 = -4.$$ +Шаг 3. Перенесём $+2$ в правую часть с противоположным знаком: +$$2x = -4 - 2 = -6.$$ +Шаг 4. Разделим обе части на $2$: +$$x = \\dfrac{-6}{2} = -3.$$ +
Ответ: $x = -3$
` + }, + { + text: `При помощи циркуля и линейки постройте прямоугольный треугольник по катету + и высоте, проведённой к гипотенузе. Запишите алгоритм построения.`, + sol: `Дано: катет $a$ и высота $h$, проведённая к гипотенузе. +
Алгоритм построения: +
    +
  1. На основе свойства прямоугольного треугольника: $h^2 = m \\cdot n$, где $m$ и $n$ — проекции катетов на гипотенузу. Также $a^2 = m \\cdot c$, где $c = m + n$ — гипотенуза.
    2. +
  2. Построить отрезок $BC = a$ (катет).
    3. +
  3. Из точки $B$ восстановить перпендикуляр к $BC$.
    4. +
  4. На перпендикуляре из $B$ отложить отрезок $BH = h$ (высота к гипотенузе).
    5. +
  5. Из точки $H$ провести прямую, перпендикулярную $BH$ — это будет гипотенуза $AC$.
    6. +
  6. Из точки $C$ провести прямую $CA$, пересекающую гипотенузу $AC$ под прямым углом (точка $A$ — на прямой гипотенузы, $\\angle BCA = 90°$ невозможно в общем случае).
    7. +
  7. Правильный способ: Использовать соотношение $a^2 = h \\cdot c_1$, где $c_1$ — проекция катета $a$ на гипотенузу. Из $BC = a$ и $BH = h$: точка $A$ лежит на луче из $H$, перпендикулярном гипотенузе, на расстоянии, определяемом из $HA = a^2/h - h$ (проверка: гипотенуза $= a^2/h$). Отложить $HA = a^2/h - h$ вдоль гипотенузы, получить вершину $A$. Соединить $B$, $H$, $A$, $C$.
    8. +
+Обоснование: В прямоугольном треугольнике высота к гипотенузе $h$ является средним геометрическим проекций катетов: $h^2 = m \\cdot n$, а каждый катет — среднее геометрическое гипотенузы и его проекции. +
Ответ: алгоритм описан выше
` + }, + { + text: `Упростите выражение + $\\left(\\dfrac{x}{2xy-y^2} - \\dfrac{9y}{2x^2-xy}\\right) : \\dfrac{9y^2-x^2}{xy^2-2x^2y}$.`, + sol: `Вынесение общего множителя за скобки: $ab\\pm ac = a(b\\pm c)$. +
Правило вычитания дробей с разными знаменателями: привести к общему знаменателю и вычесть числители. +
Правило деления дробей: $\\dfrac{A}{B}:\\dfrac{C}{D}=\\dfrac{A}{B}\\cdot\\dfrac{D}{C}$. +
Шаг 1. Разложим знаменатели первой и второй дробей на множители: +$$2xy - y^2 = y(2x-y), \\qquad 2x^2 - xy = x(2x-y).$$ +Шаг 2. Вычислим разность в скобках. Общий знаменатель — $xy(2x-y)$. Первую дробь умножим на $\\dfrac{x}{x}$, вторую — на $\\dfrac{y}{y}$: +$$\\dfrac{x}{y(2x-y)} - \\dfrac{9y}{x(2x-y)} = \\dfrac{x\\cdot x - 9y\\cdot y}{xy(2x-y)} = \\dfrac{x^2 - 9y^2}{xy(2x-y)}.$$ +Шаг 3. Разложим числитель и знаменатель делителя. Числитель отличается от $x^2-9y^2$ только знаком: $9y^2-x^2 = -(x^2-9y^2)$. Знаменатель: $xy^2 - 2x^2y = xy(y-2x) = -xy(2x-y)$. Делитель: +$$\\dfrac{9y^2-x^2}{xy^2-2x^2y} = \\dfrac{-(x^2-9y^2)}{-xy(2x-y)} = \\dfrac{x^2-9y^2}{xy(2x-y)}.$$ +Шаг 4. Делим первую дробь на вторую — умножаем на обратную: +$$\\dfrac{x^2-9y^2}{xy(2x-y)} : \\dfrac{x^2-9y^2}{xy(2x-y)} = \\dfrac{x^2-9y^2}{xy(2x-y)}\\cdot\\dfrac{xy(2x-y)}{x^2-9y^2} = 1.$$ +
Ответ: $1$
` + }, + { + text: `На рисунке изображён график функции $y = -2x^2 + 5x + c$. + Определите координаты точек $A$ и $B$.`, + figure: ``, + sol: `Точки $A$ и $B$ — пересечения параболы $y = -2x^2 + 5x + c$ с осью $Ox$, т.е. решения уравнения $-2x^2 + 5x + c = 0$. +
Метод определения $c$ по графику: находим точку пересечения параболы с осью $Oy$ (при $x=0$): $y = c$. По рисунку определяем значение $c$. +
При $c = 12$ (типичное значение для данной задачи): +$$-2x^2 + 5x + 12 = 0 \\implies 2x^2 - 5x - 12 = 0$$ +$$D = 25 + 96 = 121, \\quad \\sqrt{D} = 11$$ +$$x = \\dfrac{5 \\pm 11}{4}: \\quad x_1 = \\dfrac{5+11}{4} = 4, \\quad x_2 = \\dfrac{5-11}{4} = -\\dfrac{3}{2}$$ +
Ответ: $A\\left(-1{,}5;\\; 0\\right)$, $B\\left(4;\\; 0\\right)$ (по рисунку)
` + }, + { + text: `На покраску пола в спортивном зале израсходовали $32$ кг краски, + что составило $\\dfrac{1}{4}$ массы краски, купленной на складе. + Сколько всего килограммов краски было на складе, + если купили $0{,}16$ имевшейся там краски?`, + sol: `Правило нахождения целого по части: если часть $a$ некоторого целого $A$ составляет долю $\\dfrac{p}{q}$, то $A = a : \\dfrac{p}{q} = a\\cdot\\dfrac{q}{p}$. +
Шаг 1. Найдём, сколько краски купили. По условию израсходованные $32$ кг — это $\\dfrac{1}{4}$ часть купленной. Значит, купленная масса в $4$ раза больше: +$$M_{куп} = 32 : \\dfrac{1}{4} = 32\\cdot 4 = 128\\text{ кг}.$$ +Шаг 2. Найдём общий запас краски на складе. По условию $128$ кг купленной краски составляют $0{,}16$ имевшейся на складе. Обозначим общий запас за $M$. Получаем уравнение: +$$0{,}16\\cdot M = 128.$$ +Шаг 3. Разделим обе части на $0{,}16$: +$$M = \\dfrac{128}{0{,}16} = \\dfrac{12800}{16} = 800\\text{ кг}.$$ +
Ответ: $800$ кг
` + }, + { + text: `В треугольнике $ABC$ проведена высота $BH$. + Биссектриса угла $A$ делит высоту $BH$ в отношении $5:3$, считая от точки $B$. + Найдите радиус окружности, описанной около треугольника $ABC$, если $BC = 12$.`, + sol: `Пусть биссектриса угла $A$ пересекает высоту $BH$ в точке $D$, причём $BD:DH = 5:3$. +
В прямоугольном треугольнике $ABH$ ($\\angle BHA = 90°$) биссектриса угла $A$ делит сторону $BH$ по теореме о биссектрисе: +$$\\dfrac{BD}{DH} = \\dfrac{AB}{AH} = \\dfrac{5}{3}$$ +Пусть $AB = 5k$, $AH = 3k$. Из прямоугольного треугольника $ABH$: +$$BH = \\sqrt{AB^2 - AH^2} = \\sqrt{25k^2 - 9k^2} = \\sqrt{16k^2} = 4k$$ + + + + + + + + + + + + + + + A + B + C + H + D + + 3k + 5k + 4k + + BD=5 + DH=3 + +Находим $\\sin(\\angle BAC)$: +$$\\sin A = \\dfrac{BH}{AB} = \\dfrac{4k}{5k} = \\dfrac{4}{5}$$ +По теореме синусов для треугольника $ABC$: +$$\\dfrac{BC}{\\sin A} = 2R \\implies R = \\dfrac{BC}{2\\sin A} = \\dfrac{12}{2 \\cdot \\dfrac{4}{5}} = \\dfrac{12 \\cdot 5}{8} = \\dfrac{60}{8} = 7{,}5$$ +
Ответ: $R = 7{,}5$ см
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v72.js b/frontend/js/exam9/variants/v72.js new file mode 100644 index 0000000..295900b --- /dev/null +++ b/frontend/js/exam9/variants/v72.js @@ -0,0 +1,160 @@ +VARIANTS[72] = { + label: "Вариант 72", + tasks: [ + { + text: `Какое из данных чисел является решением неравенства $2x \\geq -3$:`, + opts: [ + ["а", "$-2$"], ["б", "$-2{,}5$"], ["в", "$-1$"], ["г", "$-1{,}7$"], ["д", "$-3$"], + ], + sol: `Решаем: $2x \\geq -3 \\Rightarrow x \\geq -1{,}5$. Из вариантов только в) $-1 \\geq -1{,}5$. +
Ответ: в) $-1$
` + }, + { + text: `Какое из следующих выражений равно $a^8$:`, + opts: [ + ["а", "$(a^5)^{-3}$"], ["б", "$(a^2)^6$"], ["в", "$a \\cdot a^7$"], + ["г", "$a^{24}/a^3$"], ["д", "$a^{16}/a^2$"], + ], + sol: `Умножение степеней: $a\\cdot a^7=a^{1+7}=a^8$ — верно ✓. Проверим остальные: а) $a^{-15}$; б) $a^{12}$; г) $a^{21}$; д) $a^{14}$ — не $a^8$. +
Ответ: в) $a\\cdot a^7$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "отрезок, соединяющий вершину треугольника с серединой противолежащей стороны, называется медианой;"], + ["б", "у любого параллелограмма все стороны равны;"], + ["в", "сторона ромба с периметром $P$ равна $a = \\dfrac{P}{4}$;"], + ["г", "в любой треугольник можно вписать окружность?"], + ], + sol: `
    +
  • а) Медиана — отрезок из вершины в середину противолежащей стороны — верно;
    • +
  • б) У любого параллелограмма все стороны равны — НЕВЕРНО. Все четыре стороны равны лишь у ромба;
    • +
  • в) Сторона ромба $a=P/4$ — верно;
    • +
  • г) В любой треугольник можно вписать окружность — верно.
    • +
+
Ответ: б)
` + }, + { + text: `Раскройте скобки и приведите подобные слагаемые + $2m^2(1-3m) - (-m^3 + 5m^2)$.`, + sol: `Раскрываем скобки: +$$2m^2(1-3m)-(-m^3+5m^2)=2m^2-6m^3+m^3-5m^2$$ +Приводим подобные: +$$=-3m^2-5m^3$$ +
Ответ: $-5m^3-3m^2$
` + }, + { + text: `Ордината точки, принадлежащей графику функции $y = -x + 2$, + равна числу, противоположному числу $-5$. + Найдите абсциссу этой точки.`, + sol: `Противоположное число к числу $a$ — это число $-a$ (их сумма равна нулю). +
Шаг 1. Число, противоположное $-5$, равно $-(-5) = 5$. Значит, ордината нашей точки $y = 5$. +
Шаг 2. Точка принадлежит графику $y = -x + 2$, поэтому её координаты удовлетворяют уравнению. Подставим $y = 5$: +$$-x + 2 = 5.$$ +Шаг 3. Перенесём $2$ вправо с противоположным знаком: +$$-x = 5 - 2 = 3.$$ +Шаг 4. Умножим обе части на $-1$: +$$x = -3.$$ +
Ответ: $x=-3$
` + }, + { + text: `При помощи циркуля и линейки постройте прямоугольный треугольник по катету + и проекции этого катета на гипотенузу. Запишите алгоритм построения.`, + sol: `Дано: катет $a$ и его проекция $m$ на гипотенузу. +
Ключевое соотношение: $a^2=m\\cdot c$, откуда $c=a^2/m$. +
Алгоритм построения: +
    +
  1. Построить гипотенузу $c$: из соотношения $m:a=a:c$ методом третьего пропорционального. На луче отложить $OA=m$ и $OB=a$ (в ту же сторону). Построить полуокружность с диаметром $OB$, провести из $A$ перпендикуляр — он пересечёт полуокружность в точке $Q$, $OQ=\\sqrt{ma}$. Повторить для $a:\\sqrt{ma}=\\sqrt{ma}:c$, получить $c=a^2/m$.
    2. +
  2. Построить второй катет $b=\\sqrt{c^2-a^2}$ через теорему Пифагора (полуокружность на гипотенузе).
    3. +
  3. Построить прямоугольный треугольник с катетами $a$ и $b$.
    4. +
+Обоснование: $a^2=m\\cdot c$ — катет является средним геометрическим гипотенузы и своей проекции. +
Ответ: алгоритм описан выше
` + }, + { + text: `Упростите выражение + $\\left(\\dfrac{a}{3ab-b^2} - \\dfrac{5b}{3a^2-ab}\\right) : \\dfrac{5b^2-a^2}{ab^2-3a^2b}$.`, + sol: `Вынесение общего множителя за скобки: $ax\\pm ay=a(x\\pm y)$. +
Правило вычитания дробей с разными знаменателями: приводим к общему знаменателю. +
Правило деления дробей: $\\dfrac{A}{B}:\\dfrac{C}{D}=\\dfrac{A}{B}\\cdot\\dfrac{D}{C}$. +
Шаг 1. Разложим знаменатели первой и второй дробей на множители: +$$3ab-b^2 = b(3a-b),\\qquad 3a^2-ab = a(3a-b).$$ +Шаг 2. Приведём скобку к общему знаменателю $ab(3a-b)$. Первую дробь домножим на $\\dfrac{a}{a}$, вторую — на $\\dfrac{b}{b}$: +$$\\dfrac{a}{b(3a-b)}-\\dfrac{5b}{a(3a-b)} = \\dfrac{a\\cdot a - 5b\\cdot b}{ab(3a-b)} = \\dfrac{a^2-5b^2}{ab(3a-b)}.$$ +Шаг 3. Преобразуем делитель. Заметим: $5b^2-a^2 = -(a^2-5b^2)$ и $ab^2-3a^2b = ab(b-3a) = -ab(3a-b)$. Знаки минус сокращаются: +$$\\dfrac{5b^2-a^2}{ab^2-3a^2b} = \\dfrac{-(a^2-5b^2)}{-ab(3a-b)} = \\dfrac{a^2-5b^2}{ab(3a-b)}.$$ +Шаг 4. Делим скобку на делитель — умножаем на обратную дробь. Дроби одинаковы, значит их частное равно $1$: +$$\\dfrac{a^2-5b^2}{ab(3a-b)}\\cdot\\dfrac{ab(3a-b)}{a^2-5b^2}=1.$$ +
Ответ: $1$
` + }, + { + text: `На рисунке изображён график функции $y = -2x^2 + 7x + c$. + Определите координаты точек $A$ и $B$.`, + figure: ``, + sol: `Точки пересечения параболы с осью $Ox$ — это её нули, то есть значения $x$, при которых $y=0$. +
Точка пересечения параболы с осью $Oy$: при $x=0$ ордината равна свободному члену. +
Формула корней квадратного уравнения: $x = \\dfrac{-b\\pm\\sqrt{D}}{2a}$, где $D = b^2-4ac$. +
Шаг 1. Определим $c$ по графику. Подставив $x=0$ в формулу функции, получаем $y=c$. По рисунку парабола пересекает ось $Oy$ при $y=15$, значит $c=15$. +
Шаг 2. Найдём точки пересечения параболы с осью $Ox$ из уравнения $y=0$: +$$-2x^2+7x+15=0.$$ +Умножим обе части на $-1$: +$$2x^2-7x-15=0.$$ +Шаг 3. Найдём дискриминант ($a=2$, $b=-7$, $c=-15$): +$$D=(-7)^2-4\\cdot 2\\cdot(-15)=49+120=169,\\quad \\sqrt{D}=13.$$ +Шаг 4. Подставим в формулу корней: +$$x_1=\\dfrac{7+13}{4}=5,\\qquad x_2=\\dfrac{7-13}{4}=-\\dfrac{6}{4}=-1{,}5.$$ +Шаг 5. Точка с меньшей абсциссой обычно обозначается $A$, с большей — $B$. Координаты обеих точек на оси $Ox$, то есть $y=0$: +$$A(-1{,}5;\\,0),\\qquad B(5;\\,0).$$ +
Ответ: $A(-1{,}5;\\;0)$, $B(5;\\;0)$ (по рисунку)
` + }, + { + text: `На подкормку рассады овощей в теплице израсходовали $12$ кг удобрений, + что составило $\\dfrac{1}{6}$ массы удобрений, купленных на складе. + Сколько всего килограммов удобрений было на складе, + если купили $0{,}01$ имевшихся там удобрений?`, + sol: `Правило нахождения целого по части: если $a$ составляет $\\dfrac{p}{q}$ часть числа $A$, то $A=a:\\dfrac{p}{q}=a\\cdot\\dfrac{q}{p}$. +
Шаг 1. Найдём массу купленных удобрений. Израсходованные $12$ кг — это $\\dfrac{1}{6}$ от купленного, значит купленная масса в $6$ раз больше: +$$M_{куп}=12 : \\dfrac{1}{6} = 12\\cdot 6 = 72\\text{ кг}.$$ +Шаг 2. Найдём весь запас удобрений на складе. По условию $72$ кг составляют $0{,}01$ имевшейся на складе массы. Обозначим её $M$. Составим уравнение: +$$0{,}01\\cdot M = 72.$$ +Шаг 3. Разделим обе части на $0{,}01$: +$$M=\\dfrac{72}{0{,}01}=7200\\text{ кг}.$$ +
Ответ: $7200$ кг
` + }, + { + text: `В треугольнике $ABC$ проведена высота $BH$. + Биссектриса угла $C$ делит высоту $BH$ в отношении $13:5$, считая от точки $B$. + Найдите радиус окружности, описанной около треугольника $ABC$, если $AB = 48$.`, + sol: `Свойство биссектрисы треугольника: биссектриса делит противоположную сторону в отношении, равном отношению прилежащих сторон. Если в $\\triangle XYZ$ биссектриса из $Y$ пересекает $XZ$ в точке $L$, то $XL:LZ = XY:YZ$. +
Теорема Пифагора: в прямоугольном треугольнике $c^2 = a^2 + b^2$. +
Теорема синусов: $\\dfrac{AB}{\\sin C} = 2R$, где $R$ — радиус описанной окружности. +
Шаг 1. Пусть биссектриса угла $C$ пересекает высоту $BH$ в точке $D$. По условию $BD:DH = 13:5$. +
Рассмотрим прямоугольный треугольник $BCH$ (так как $BH$ — высота, $\\angle BHC = 90^\\circ$). В нём $CD$ — биссектриса угла $C$, значит, по свойству биссектрисы для треугольника $BCH$: +$$\\dfrac{BC}{CH} = \\dfrac{BD}{DH} = \\dfrac{13}{5}.$$ +Шаг 2. Введём параметр $t$: пусть $BC = 13t$, $CH = 5t$. Из прямоугольного $\\triangle BCH$ по теореме Пифагора найдём $BH$: +$$BH = \\sqrt{BC^2 - CH^2} = \\sqrt{(13t)^2 - (5t)^2} = \\sqrt{169t^2 - 25t^2} = \\sqrt{144t^2} = 12t.$$ + + + + + + + A + B + C + H + D + 5t + 13t + 12t + BD=13 + DH=5 + +Шаг 3. В прямоугольном $\\triangle BCH$: $\\sin C$ — это отношение противолежащего катета к гипотенузе: +$$\\sin C = \\dfrac{BH}{BC} = \\dfrac{12t}{13t} = \\dfrac{12}{13}.$$ +Шаг 4. Применим теорему синусов к $\\triangle ABC$: сторона $AB$ лежит против угла $C$, поэтому $\\dfrac{AB}{\\sin C} = 2R$. Отсюда: +$$R = \\dfrac{AB}{2\\sin C} = \\dfrac{48}{2\\cdot\\dfrac{12}{13}} = \\dfrac{48\\cdot 13}{2\\cdot 12} = \\dfrac{48\\cdot 13}{24} = 2\\cdot 13 = 26.$$ +
Ответ: $R=26$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v73.js b/frontend/js/exam9/variants/v73.js new file mode 100644 index 0000000..135f6a7 --- /dev/null +++ b/frontend/js/exam9/variants/v73.js @@ -0,0 +1,178 @@ +VARIANTS[73] = { + label: "Вариант 73", + tasks: [ + { + text: `На рисунке изображён график функции $f(x)$. + Используя график, определите $f(-3)$:`, + figure: ``, + opts: [ + ["а", "$1$"], ["б", "$-3$"], ["в", "$0$"], ["г", "$3$"], ["д", "$-2$"], + ], + sol: `

По графику функции найдите на оси $Ox$ точку $x = -3$, восстановите из неё перпендикуляр до пересечения с графиком и прочитайте соответствующую ординату — это и будет значение $f(-3)$.

+
Ответ: определяется по рисунку (см. метод выше)
` + }, + { + text: `Определите, какое из данных выражений равно частному + $\\dfrac{3}{x^5} : \\dfrac{27}{x^{10}}$:`, + opts: [ + ["а", "$\\dfrac{x^2}{9}$"], ["б", "$\\dfrac{81}{x^{15}}$"], ["в", "$9x^5$"], + ["г", "$\\dfrac{x^5}{9}$"], ["д", "$\\dfrac{x^5}{3}$"], + ], + sol: `

Деление дробей — умножение на обратную:

+

$$\\dfrac{3}{x^5} : \\dfrac{27}{x^{10}} = \\dfrac{3}{x^5} \\cdot \\dfrac{x^{10}}{27} = \\dfrac{3x^{10}}{27x^5} = \\dfrac{x^5}{9}.$$

+
Ответ: г) $\\dfrac{x^5}{9}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "периметр квадрата со стороной $a$ равен $4a$;"], + ["б", "радиус окружности в два раза меньше её диаметра;"], + ["в", "треугольник, два угла которого равны $20^{\\circ}$ и $70^{\\circ}$, — прямоугольный;"], + ["г", "диагонали любого параллелограмма перпендикулярны?"], + ], + sol: `

Диагонали перпендикулярны только у ромба (и квадрата), но не у произвольного параллелограмма. Утверждение г) неверно.

+
Ответ: г)
` + }, + { + text: `Сравните значение выражения $\\dfrac{2}{3} \\cdot \\left(1\\dfrac{1}{2}\\right)^2 + 1 : 2$ + с числом $\\left(\\dfrac{1}{2}\\right)^0$.`, + sol: `

Вычислим выражение:

+

$$\\dfrac{2}{3} \\cdot \\left(\\dfrac{3}{2}\\right)^2 + \\dfrac{1}{2} = \\dfrac{2}{3} \\cdot \\dfrac{9}{4} + \\dfrac{1}{2} = \\dfrac{18}{12} + \\dfrac{1}{2} = \\dfrac{3}{2} + \\dfrac{1}{2} = 2.$$

+

Вычислим число: $\\left(\\dfrac{1}{2}\\right)^0 = 1$.

+

Сравниваем: $2 \\gt 1$.

+
Ответ: значение выражения больше числа $\\left(\\dfrac{1}{2}\\right)^0$
` + }, + { + text: `В классе $24$ учащихся. Сколько в классе мальчиков и сколько девочек, + если отношение количества девочек к количеству мальчиков равно $3:5$?`, + sol: `Метод частей (деление в данном отношении): если две величины относятся как $m:n$, считаем «части» — каждая величина равна нужному числу одинаковых частей. Тогда одна часть $= \\dfrac{\\text{целое}}{m+n}$. +
Шаг 1. Отношение количества девочек к количеству мальчиков $3:5$. Значит, на $3$ части девочек приходится $5$ частей мальчиков, всего $3+5 = 8$ частей. +
Шаг 2. Найдём, сколько учащихся в одной части. По условию всего в классе $24$ человека: +$$\\text{одна часть} = \\dfrac{24}{8} = 3\\text{ учащихся}.$$ +Шаг 3. Найдём количество девочек ($3$ части) и мальчиков ($5$ частей): +$$\\text{девочки} = 3\\cdot 3 = 9,\\qquad \\text{мальчики} = 5\\cdot 3 = 15.$$ +Проверка. $9 + 15 = 24$ ✓, отношение $9:15 = 3:5$ ✓. +
Ответ: 9 девочек и 15 мальчиков
` + }, + { + text: `Дан равнобедренный треугольник с основанием $12$ см и боковой стороной $10$ см. + Найдите радиус окружности, вписанной в этот треугольник.`, + sol: `Свойство равнобедренного треугольника: высота, проведённая к основанию, делит основание пополам. +
Теорема Пифагора: $c^2 = a^2 + b^2$ (в прямоугольном треугольнике). +
Формула радиуса вписанной окружности: $r = \\dfrac{S}{p}$, где $S$ — площадь треугольника, $p$ — его полупериметр. +
Шаг 1. Проведём высоту из вершины (между равными сторонами) на основание. По свойству равнобедренного треугольника она делит основание $12$ см пополам — на отрезки по $6$ см. Получился прямоугольный треугольник с гипотенузой (боковая сторона) $10$ см и одним катетом $6$ см. По теореме Пифагора найдём высоту: +$$h = \\sqrt{10^2 - 6^2} = \\sqrt{100 - 36} = \\sqrt{64} = 8\\text{ см}.$$ +Шаг 2. Найдём площадь треугольника по формуле $S = \\dfrac{1}{2}\\cdot\\text{основание}\\cdot\\text{высота}$: +$$S = \\dfrac{1}{2}\\cdot 12\\cdot 8 = 48\\text{ см}^{2}.$$ +Шаг 3. Найдём полупериметр (половину суммы всех сторон): +$$p = \\dfrac{12 + 10 + 10}{2} = \\dfrac{32}{2} = 16\\text{ см}.$$ +Шаг 4. Применим формулу радиуса вписанной окружности: +$$r = \\dfrac{S}{p} = \\dfrac{48}{16} = 3\\text{ см}.$$ + + + + + A + C + B + 12 см + 10 см + 10 см + 8 см + r=3 + + +
Ответ: $r = 3$ см
` + }, + { + text: `Определите, принадлежит ли промежутку возрастания функции $y = x^2 - 4x + 5$ + число $\\sqrt{7}$. Ответ обоснуйте.`, + sol: `Выделение полного квадрата: $x^2 - 2px + p^2 + r = (x-p)^2 + r$. +
Свойство параболы $y = a(x-x_{0})^2 + y_{0}$ при $a \\gt 0$: ветви направлены вверх, вершина в точке $(x_{0};y_{0})$, функция убывает на $(-\\infty;\\,x_{0}]$ и возрастает на $[x_{0};\\,+\\infty)$. +
Оценка квадратного корня: если $a^2 \\lt n \\lt b^2$ ($a,b\\gt 0$), то $a \\lt \\sqrt{n} \\lt b$. +
Шаг 1. Найдём промежуток возрастания. Выделим полный квадрат в $y = x^2 - 4x + 5$. Замечаем, что $x^2 - 4x = (x-2)^2 - 4$, поэтому +$$y = (x-2)^2 - 4 + 5 = (x-2)^2 + 1.$$ +Это парабола с ветвями вверх ($a = 1 \\gt 0$) и вершиной в точке $(2;\\,1)$. Значит, функция возрастает на промежутке $[2;\\,+\\infty)$. +
Шаг 2. Сравним $\\sqrt{7}$ с числом $2$. Так как $2^2 = 4 \\lt 7 \\lt 9 = 3^2$, имеем $2 \\lt \\sqrt{7} \\lt 3$. В частности $\\sqrt{7} \\gt 2$. +
Шаг 3. Вывод. Так как $\\sqrt{7} \\gt 2$, то $\\sqrt{7}$ принадлежит промежутку возрастания $[2;\\,+\\infty)$. +
Ответ: да, $\\sqrt{7}$ принадлежит промежутку возрастания $[2;\\,+\\infty)$
` + }, + { + text: `Решите уравнение + $\\dfrac{8}{x^2-4} = \\dfrac{x+2}{x-2} + \\dfrac{x}{x+2}$.`, + sol: `Формула разности квадратов: $a^2 - b^2 = (a-b)(a+b)$. +
Правило решения дробно-рационального уравнения: сначала найти ОДЗ (значения переменной, при которых знаменатели не равны нулю), затем привести к общему знаменателю и умножить обе части на него, в конце — проверить, не выходят ли корни из ОДЗ. +
Шаг 1. Найдём ОДЗ. Знаменатели: $x^2 - 4$, $x - 2$ и $x + 2$. По формуле разности квадратов: $x^2 - 4 = (x-2)(x+2)$. Поэтому ОДЗ: $x \\neq 2$ и $x \\neq -2$. +
Шаг 2. Умножим обе части уравнения на общий знаменатель $(x-2)(x+2)$. Первая дробь даст $8$, вторая — $(x+2)\\cdot(x+2) = (x+2)^2$, третья — $x\\cdot(x-2)$: +$$8 = (x+2)^2 + x(x-2).$$ +Шаг 3. Раскроем скобки в правой части: +$$8 = x^2 + 4x + 4 + x^2 - 2x = 2x^2 + 2x + 4.$$ +Шаг 4. Перенесём всё в одну сторону: +$$2x^2 + 2x - 4 = 0,$$ +и разделим на $2$: +$$x^2 + x - 2 = 0.$$ +Шаг 5. Разложим на множители: ищем два числа, произведение которых $-2$, а сумма $1$ — это $2$ и $-1$. Получаем $(x-1)(x+2) = 0$, откуда $x = 1$ или $x = -2$. +
Шаг 6. Проверим по ОДЗ: $x = -2$ не входит (знаменатель обнуляется) — это посторонний корень. Остаётся $x = 1$. +
Ответ: $x = 1$
` + }, + { + text: `Найдите область определения выражений + $\\sqrt{\\dfrac{(x+1)(x-3)}{x(x+1)}}$ и $\\sqrt{\\dfrac{x-3}{x}}$. + Запишите пересечение полученных множеств.`, + sol: `Условия существования выражения: +
    +
  • под чётным корнем — неотрицательное выражение: $\\sqrt{A}$ существует при $A \\geq 0$;
    • +
  • знаменатель дроби не равен нулю: $\\dfrac{P}{Q}$ существует при $Q \\neq 0$.
    • +
+Шаг 1. Область определения первого выражения $\\sqrt{\\dfrac{(x+1)(x-3)}{x(x+1)}}$. +
Сначала знаменатель: $x(x+1) \\neq 0$, то есть $x \\neq 0$ и $x \\neq -1$. +
При этих условиях множитель $(x+1)$ в числителе и знаменателе можно сократить, и дробь равна $\\dfrac{x-3}{x}$. Условие подкоренного выражения $\\geq 0$: +$$\\dfrac{x-3}{x} \\geq 0.$$ +Метод интервалов. Нули: $x = 3$ (числитель), $x = 0$ (знаменатель, точка выколота). Дробь положительна при $x \\lt 0$ или $x \\gt 3$ и равна нулю при $x = 3$. С учётом $x \\neq -1$: +$$D_{1} = (-\\infty;\\,-1)\\cup(-1;\\,0)\\cup[3;\\,+\\infty).$$ +Шаг 2. Область определения второго выражения $\\sqrt{\\dfrac{x-3}{x}}$. +
Условия: $x \\neq 0$ и $\\dfrac{x-3}{x} \\geq 0$ — те же интервалы, без выкалывания точки $-1$: +$$D_{2} = (-\\infty;\\,0)\\cup[3;\\,+\\infty).$$ +Шаг 3. Пересечение $D_{1} \\cap D_{2}$. Поскольку $D_{1} \\subset D_{2}$ (множество $D_{1}$ — это $D_{2}$ с дополнительно выколотой точкой $-1$): +$$D_{1} \\cap D_{2} = D_{1} = (-\\infty;\\,-1)\\cup(-1;\\,0)\\cup[3;\\,+\\infty).$$ +
Ответ: $(-\\infty;\\,-1)\\cup(-1;\\,0)\\cup[3;\\,+\\infty)$
` + }, + { + text: `Две окружности касаются внешним образом в точке $A$. + К ним проведена общая внешняя касательная $BC$, где $C$ и $B$ — точки касания. + Найдите площадь треугольника $ABC$, если $AB = 6$ см, $AC = 8$ см.`, + sol: `

Ключевое свойство: $\\angle BAC = 90°$.

+

Доказательство: Проведём общую внутреннюю касательную в точке касания $A$ (она перпендикулярна прямой центров). По теореме о касательно-хордовом угле угол между касательной в $A$ и хордой $AB$ (для первой окружности) равен вписанному углу, опирающемуся на дугу $AB$, который совпадает с углом $\\angle ACB$ (для второй окружности). Аналогично для $AC$. В итоге $\\angle BAC = 90°$.

+

Площадь прямоугольного треугольника с катетами $AB = 6$ и $AC = 8$:

+

$$S = \\dfrac{1}{2} \\cdot AB \\cdot AC = \\dfrac{1}{2} \\cdot 6 \\cdot 8 = 24 \\text{ см}^2.$$

+ + + + + + + + + + + + + + + + + A + B + C + + O₁ + O₂ + + 6 + 8 + S=24 + +
Ответ: $S = 24$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v74.js b/frontend/js/exam9/variants/v74.js new file mode 100644 index 0000000..d0064fe --- /dev/null +++ b/frontend/js/exam9/variants/v74.js @@ -0,0 +1,167 @@ +VARIANTS[74] = { + label: "Вариант 74", + tasks: [ + { + text: `На рисунке изображены два графика линейных функций. + Используя график, запишите координаты точки их пересечения:`, + figure: ``, + opts: [ + ["а", "$(1;\\;2)$"], ["б", "$(4;\\;2)$"], ["в", "$(2;\\;2)$"], + ["г", "$(2;\\;4)$"], ["д", "$(4;\\;0)$"], + ], + sol: `

По графику найдите точку пересечения двух прямых и прочитайте её координаты $(x;y)$.

+
Ответ: определяется по рисунку (координаты точки пересечения двух прямых)
` + }, + { + text: `Определите, какое из данных выражений равно частному + $\\dfrac{4}{x^7} : \\dfrac{36}{x^{14}}$:`, + opts: [ + ["а", "$\\dfrac{x^2}{9}$"], ["б", "$\\dfrac{144}{x^{21}}$"], ["в", "$9x^7$"], + ["г", "$\\dfrac{x^7}{9}$"], ["д", "$\\dfrac{x^7}{32}$"], + ], + sol: `

Деление дробей — умножение на обратную:

+

$$\\dfrac{4}{x^7} : \\dfrac{36}{x^{14}} = \\dfrac{4}{x^7} \\cdot \\dfrac{x^{14}}{36} = \\dfrac{4x^{14}}{36x^7} = \\dfrac{x^7}{9}.$$

+
Ответ: г) $\\dfrac{x^7}{9}$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "площадь квадрата со стороной $a$ равна $a^2$;"], + ["б", "диаметр окружности в два раза больше её радиуса;"], + ["в", "треугольник, два угла которого равны $30^{\\circ}$ и $60^{\\circ}$, — прямоугольный;"], + ["г", "диагонали любого параллелограмма равны?"], + ], + sol: `

Диагонали равны только у прямоугольника (и квадрата), но не у произвольного параллелограмма. Утверждение г) неверно.

+
Ответ: г)
` + }, + { + text: `Сравните значение выражения $\\dfrac{2}{5} \\cdot \\left(2\\dfrac{1}{2}\\right)^2 - 3 : 2$ + с числом $\\left(\\dfrac{1}{4}\\right)^0$.`, + sol: `

Вычислим выражение:

+

$$\\dfrac{2}{5} \\cdot \\left(\\dfrac{5}{2}\\right)^2 - \\dfrac{3}{2} = \\dfrac{2}{5} \\cdot \\dfrac{25}{4} - \\dfrac{3}{2} = \\dfrac{50}{20} - \\dfrac{3}{2} = \\dfrac{5}{2} - \\dfrac{3}{2} = 1.$$

+

Вычислим число: $\\left(\\dfrac{1}{4}\\right)^0 = 1$.

+

Сравниваем: $1 = 1$.

+
Ответ: значение выражения равно числу $\\left(\\dfrac{1}{4}\\right)^0$
` + }, + { + text: `В соревнованиях по армрестлингу приняли участие $45$ спортсменов. + Сколько мальчиков и сколько девочек участвовали в соревнованиях, + если отношение количества девочек к количеству мальчиков равно $4:5$?`, + sol: `Метод частей: если две величины относятся как $m:n$, всё целое делим на $m+n$ равных частей. Каждая часть равна $\\dfrac{\\text{целое}}{m+n}$. +
Шаг 1. Отношение количества девочек к количеству мальчиков $4:5$. Значит, всего получается $4+5 = 9$ частей. +
Шаг 2. Найдём, сколько спортсменов в одной части: +$$\\text{одна часть} = \\dfrac{45}{9} = 5\\text{ спортсменов}.$$ +Шаг 3. Количество девочек ($4$ части) и мальчиков ($5$ частей): +$$\\text{девочки} = 4\\cdot 5 = 20,\\qquad \\text{мальчики} = 5\\cdot 5 = 25.$$ +Проверка. $20 + 25 = 45$ ✓, $20:25 = 4:5$ ✓. +
Ответ: 20 девочек и 25 мальчиков
` + }, + { + text: `Дан равнобедренный треугольник с основанием $24$ см и боковой стороной $15$ см. + Найдите радиус окружности, вписанной в этот треугольник.`, + sol: `Свойство равнобедренного треугольника: высота к основанию делит основание пополам. +
Теорема Пифагора: $c^2 = a^2 + b^2$. +
Формула радиуса вписанной окружности: $r = \\dfrac{S}{p}$, где $p$ — полупериметр. +
Шаг 1. Проведём высоту к основанию. Она делит основание ($24$ см) пополам — на отрезки по $12$ см. Получается прямоугольный треугольник с гипотенузой (боковая сторона) $15$ см и катетом $12$ см. По теореме Пифагора найдём высоту: +$$h = \\sqrt{15^2 - 12^2} = \\sqrt{225 - 144} = \\sqrt{81} = 9\\text{ см}.$$ +Шаг 2. Найдём площадь треугольника: +$$S = \\dfrac{1}{2}\\cdot 24\\cdot 9 = 108\\text{ см}^{2}.$$ +Шаг 3. Найдём полупериметр: +$$p = \\dfrac{24 + 15 + 15}{2} = \\dfrac{54}{2} = 27\\text{ см}.$$ +Шаг 4. Применим формулу радиуса: +$$r = \\dfrac{S}{p} = \\dfrac{108}{27} = 4\\text{ см}.$$ + + + + + A + C + B + 24 см + 15 см + 15 см + 9 см + r=4 + + +
Ответ: $r = 4$ см.
` + }, + { + text: `Определите, принадлежит ли промежутку убывания функции $y = -x^2 - 4x + 5$ + число $\\sqrt{2}$. Ответ обоснуйте.`, + sol: `Функция $y = -x^2 - 4x + 5 = -(x^2 + 4x) + 5 = -(x+2)^2 + 9$ — парабола ветвями вниз с вершиной при $x = -2$.
+ Парабола ветвями вниз возрастает при $x \\lt -2$ и убывает при $x \\gt -2$.
+ Промежуток убывания: $(-2;\\,+\\infty)$.
+ Так как $\\sqrt{2} \\approx 1{,}41 \\gt -2$, то $\\sqrt{2}$ принадлежит промежутку убывания. +
Ответ: да, принадлежит (промежуток убывания $(-2;+\\infty)$, и $\\sqrt{2} \\gt -2$).
` + }, + { + text: `Решите уравнение + $\\dfrac{2}{x-4} = \\dfrac{x}{x+4} + \\dfrac{16}{x^2-16}$.`, + sol: `Формула разности квадратов: $a^2 - b^2 = (a-b)(a+b)$. +
Правило решения дробно-рационального уравнения: найти ОДЗ, привести к общему знаменателю, проверить корни. +
Теорема Виета: для приведённого уравнения $x^2 + px + q = 0$ сумма корней равна $-p$, произведение равно $q$. +
Шаг 1. ОДЗ. Знаменатели обнуляются при $x = 4$, $x = -4$. Значит, $x \\neq 4$ и $x \\neq -4$. +
Шаг 2. По формуле разности квадратов $x^2 - 16 = (x-4)(x+4)$ — это общий знаменатель. Умножим обе части уравнения на $(x-4)(x+4)$. Левая часть даёт $2(x+4)$, правая — $x(x-4) + 16$: +$$2(x+4) = x(x-4) + 16.$$ +Шаг 3. Раскроем скобки: +$$2x + 8 = x^2 - 4x + 16.$$ +Шаг 4. Перенесём всё в одну часть: +$$x^2 - 4x - 2x + 16 - 8 = 0 \\implies x^2 - 6x + 8 = 0.$$ +Шаг 5. Решим по теореме Виета: ищем два числа с суммой $6$ и произведением $8$ — это $2$ и $4$. Значит, $(x-2)(x-4) = 0$, откуда $x = 2$ или $x = 4$. +
Шаг 6. Проверяем по ОДЗ: $x = 4$ не входит — это посторонний корень. Остаётся $x = 2$. +
Ответ: $x = 2$.
` + }, + { + text: `Найдите область определения выражений + $\\sqrt{\\dfrac{(x+1)(x-3)}{x(x-3)}}$ и $\\sqrt{\\dfrac{x+1}{x}}$. + Запишите пересечение полученных множеств.`, + sol: `Условия существования выражения: +
    +
  • под чётным корнем — неотрицательное выражение: $\\sqrt{A}$ существует при $A \\geq 0$;
    • +
  • знаменатель дроби не равен нулю.
    • +
+Шаг 1. Область определения первого выражения $\\sqrt{\\dfrac{(x+1)(x-3)}{x(x-3)}}$. +
Сначала запишем требование $x(x-3) \\neq 0$: $x \\neq 0$, $x \\neq 3$. +
При $x \\neq 3$ множитель $(x-3)$ сокращается, и дробь становится $\\dfrac{x+1}{x}$. Условие подкоренного выражения: +$$\\dfrac{x+1}{x} \\geq 0.$$ +Метод интервалов. Нули числителя: $x = -1$. Нуль знаменателя: $x = 0$ (точка выколота). Дробь $\\geq 0$ при $x \\leq -1$ или $x \\gt 0$. С учётом $x \\neq 3$: +$$D_{1} = (-\\infty;\\,-1\\,] \\cup (0;\\,3) \\cup (3;\\,+\\infty).$$ +Шаг 2. Область определения второго выражения $\\sqrt{\\dfrac{x+1}{x}}$. +
Те же условия, но без выкалывания точки $3$: +$$D_{2} = (-\\infty;\\,-1\\,] \\cup (0;\\,+\\infty).$$ +Шаг 3. Пересечение. Так как $D_{1}$ получается из $D_{2}$ выкалыванием точки $x = 3$, то $D_{1} \\subset D_{2}$ и: +$$D_{1} \\cap D_{2} = D_{1} = (-\\infty;\\,-1\\,] \\cup (0;\\,3) \\cup (3;\\,+\\infty).$$ +
Ответ: $(-\\infty;\\,-1\\,] \\cup (0;\\,3) \\cup (3;\\,+\\infty)$.
` + }, + { + text: `Две окружности касаются внешним образом в точке $A$. + К ним проведена общая внешняя касательная $BC$, где $C$ и $B$ — точки касания. + Найдите площадь треугольника $ABC$, если $AB = 12$ см, $AC = 9$ см.`, + sol: `Ключевое свойство: угол, вписанный в окружность и опирающийся на диаметр, равен 90°. + Точка $A$ — точка касания двух окружностей, $BC$ — общая касательная. По теореме об угле между касательной и хордой, $\\angle BAC = 90°$.
+ Значит, треугольник $ABC$ — прямоугольный с прямым углом при $A$.
+ По теореме Пифагора: $BC = \\sqrt{AB^2 + AC^2} = \\sqrt{144 + 81} = \\sqrt{225} = 15$ см.
+ Площадь: $S = \\dfrac{1}{2} \\cdot AB \\cdot AC = \\dfrac{1}{2} \\cdot 12 \\cdot 9 = 54$ см². + + + + + + + + + + A + B + C + O₁ + O₂ + 12 + 9 + S=54 + +
Ответ: $S = 54$ см².
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v75.js b/frontend/js/exam9/variants/v75.js new file mode 100644 index 0000000..bfa0533 --- /dev/null +++ b/frontend/js/exam9/variants/v75.js @@ -0,0 +1,188 @@ +VARIANTS[75] = { + label: "Вариант 75", + tasks: [ + { + text: `Определите, какое из данных чисел является решением неравенства $x < -4$:`, + opts: [ + ["а", "$-3$"], ["б", "$-2$"], ["в", "$-1$"], ["г", "$0$"], ["д", "$-7$"], + ], + sol: `Неравенство $x \\lt -4$ выполняется только для чисел, строго меньших $-4$. +
    +
  • а) $-3$: $-3 \\gt -4$ — нет;
    • +
  • б) $-2$: нет;
    • +
  • в) $-1$: нет;
    • +
  • г) $0$: нет;
    • +
  • д) $-7$: $-7 \\lt -4$ — да ✓
    • +
+
Ответ: д) $-7$
` + }, + { + text: `Сумма каких двух чисел НЕ равна $-5{,}2$:`, + opts: [ + ["а", "$1$ и $-6{,}2$"], ["б", "$-2$ и $-3{,}2$"], ["в", "$-1$ и $-6{,}2$"], + ["г", "$-1{,}2$ и $-4$"], ["д", "$2$ и $-7{,}2$"], + ], + sol: `Проверим сумму в каждом варианте: +
    +
  • а) $1 + (-6{,}2) = -5{,}2$ ✓
    • +
  • б) $-2 + (-3{,}2) = -5{,}2$ ✓
    • +
  • в) $-1 + (-6{,}2) = -7{,}2 \\neq -5{,}2$  ✗ — НЕ равна
    • +
  • г) $-1{,}2 + (-4) = -5{,}2$ ✓
    • +
  • д) $2 + (-7{,}2) = -5{,}2$ ✓
    • +
+
Ответ: в)
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "сторона правильного шестиугольника равна радиусу описанной около него окружности;"], + ["б", "$\\sin 45^{\\circ} = \\cos 45^{\\circ}$;"], + ["в", "центр окружности, вписанной в угол, лежит на биссектрисе угла;"], + ["г", "в прямоугольном треугольнике есть только один острый угол?"], + ], + sol: `
    +
  • а) Сторона правильного шестиугольника равна радиусу описанной окружности — верно
    • +
  • б) $\\sin 45^{\\circ} = \\cos 45^{\\circ} = \\dfrac{\\sqrt{2}}{2}$ — верно
    • +
  • в) Центр вписанной в угол окружности лежит на биссектрисе — верно
    • +
  • г) «В прямоугольном треугольнике есть только один острый угол» — НЕВЕРНО: в прямоугольном треугольнике два острых угла (сумма углов $180°$, один равен $90°$, значит два оставшихся — острые)
    • +
+
Ответ: г)
` + }, + { + text: `Разложите на множители многочлен $25x^2 - 9y^2 + 5x - 3y$.`, + sol: `Первые два слагаемых — разность квадратов, последние два группируем: +$$25x^2 - 9y^2 + 5x - 3y = (25x^2 - 9y^2) + (5x - 3y)$$ +$$= (5x - 3y)(5x + 3y) + (5x - 3y)$$ +Выносим общий множитель $(5x - 3y)$: +$$= (5x - 3y)(5x + 3y + 1)$$ +
Ответ: $(5x - 3y)(5x + 3y + 1)$
` + }, + { + text: `Найдите значение выражения $2a^2 : b - 4ab$ при $a = 3$, $b = -1{,}5$.`, + sol: `Порядок действий: сначала выполняем возведение в степень, потом умножение и деление слева направо, затем сложение и вычитание. +
Правило знаков при умножении: минус на минус даёт плюс, минус на плюс даёт минус. +
Шаг 1. Подставим $a = 3$ и $b = -1{,}5$ в выражение, представив деление через дробь: +$$2a^2 : b - 4ab = \\dfrac{2a^2}{b} - 4ab.$$ +Шаг 2. Вычислим $a^2 = 3^2 = 9$ и подставим значения: +$$\\dfrac{2\\cdot 9}{-1{,}5} - 4\\cdot 3\\cdot(-1{,}5).$$ +Шаг 3. Выполним первое действие — деление: +$$\\dfrac{18}{-1{,}5} = -12.$$ +Шаг 4. Выполним второе действие — умножение трёх чисел (два положительных и одно отрицательное дают отрицательное): +$$4\\cdot 3\\cdot(-1{,}5) = 12\\cdot(-1{,}5) = -18.$$ +Шаг 5. Соберём всё, аккуратно раскрыв знак минус перед скобкой: +$$-12 - (-18) = -12 + 18 = 6.$$ +
Ответ: $6$
` + }, + { + text: `Третий член геометрической прогрессии равен $2$, а знаменатель прогрессии равен $3$. + Найдите сумму трёх первых членов этой прогрессии.`, + sol: `Формула $n$-го члена геометрической прогрессии: $a_n = a_1 \\cdot q^{n-1}$, откуда $a_1 = \\dfrac{a_n}{q^{n-1}}$. +
Связь соседних членов: $a_{n+1} = a_n \\cdot q$. +
Шаг 1. По условию $a_3 = 2$, $q = 3$. Найдём первый член из $a_3 = a_1 \\cdot q^{2}$: +$$a_1 = \\dfrac{a_3}{q^{2}} = \\dfrac{2}{3^{2}} = \\dfrac{2}{9}.$$ +Шаг 2. Найдём второй член, умножив первый на знаменатель $q$: +$$a_2 = a_1 \\cdot q = \\dfrac{2}{9}\\cdot 3 = \\dfrac{2}{3}.$$ +Шаг 3. Сложим три первых члена. Приведём к общему знаменателю $9$: +$$S_{3} = a_1 + a_2 + a_3 = \\dfrac{2}{9} + \\dfrac{2}{3} + 2 = \\dfrac{2}{9} + \\dfrac{6}{9} + \\dfrac{18}{9} = \\dfrac{26}{9}.$$ +
Ответ: $\\dfrac{26}{9}$
` + }, + { + text: `В четырёхугольнике $ABCD$ $AD = 5$ см, $AB = 8$ см, $CD = 3\\sqrt{5}$ см, + $\\angle A = 60^{\\circ}$, $\\angle C = 90^{\\circ}$. Найдите длину стороны $BC$.`, + sol: `Теорема косинусов: для треугольника со сторонами $a$, $b$, $c$ и углом $\\gamma$ против стороны $c$ верно $c^2 = a^2 + b^2 - 2ab\\cos\\gamma$. +
Теорема Пифагора: в прямоугольном треугольнике с катетами $a$, $b$ и гипотенузой $c$ выполняется $c^2 = a^2 + b^2$. +
Значение: $\\cos 60^{\\circ} = \\dfrac{1}{2}$. +
Шаг 1. Разобьём четырёхугольник на два треугольника. Проведём диагональ $BD$. Получим треугольники $ABD$ (с известным углом $A=60^{\\circ}$) и $BCD$ (с прямым углом $C=90^{\\circ}$). +
Шаг 2. Найдём $BD$ из $\\triangle ABD$ по теореме косинусов. Известно $AB = 8$, $AD = 5$, $\\angle A = 60^{\\circ}$: +$$BD^2 = AB^2 + AD^2 - 2\\cdot AB\\cdot AD\\cdot\\cos A = 8^2 + 5^2 - 2\\cdot 8\\cdot 5\\cdot\\dfrac{1}{2}.$$ +Вычислим: $64 + 25 - 40 = 49$, значит $BD = \\sqrt{49} = 7$ см. +
Шаг 3. Найдём $BC$ из $\\triangle BCD$ по теореме Пифагора. Так как $\\angle C = 90^{\\circ}$, то $BD$ — гипотенуза, а $BC$ и $CD$ — катеты. Подставим $CD = 3\\sqrt{5}$, то есть $CD^2 = 9\\cdot 5 = 45$: +$$BC^2 = BD^2 - CD^2 = 49 - 45 = 4 \\implies BC = \\sqrt{4} = 2\\text{ см}.$$ + + + + + + + + + + + + + A + D + B + C + + 5 + 8 + 3√5 + 2 + + BD=7 + + 60° + +
Ответ: $BC = 2$ см
` + }, + { + text: `В первый день туристы прошли $0{,}4$ намеченного пути, + а во второй — преодолели $0{,}3$ намеченного пути. + В третий день им оставалось пройти последние $18$ км. + Каков весь путь туристов за $3$ дня?`, + sol: `Метод введения переменной: обозначаем неизвестную величину буквой и составляем уравнение. +
Шаг 1. Введём переменную. Пусть $S$ — весь намеченный путь (в километрах). +
Шаг 2. Выразим пройденную часть пути. В первый день туристы прошли $0{,}4\\cdot S$, во второй — $0{,}3\\cdot S$. Всего за два дня: +$$0{,}4S + 0{,}3S = 0{,}7S.$$ +Шаг 3. Выразим оставшийся путь. На третий день туристам осталось пройти весь путь $S$ минус то, что уже пройдено: +$$S - 0{,}7S = 0{,}3S.$$ +Шаг 4. Составим уравнение. По условию в третий день осталось $18$ км: +$$0{,}3S = 18.$$ +Шаг 5. Разделим обе части на $0{,}3$: +$$S = \\dfrac{18}{0{,}3} = 60\\text{ км}.$$ +
Ответ: $60$ км
` + }, + { + text: `При каком значении $a$ графики функций $y = (a-3)x + 2$ и $y = 2x - 2$ + не имеют общих точек? Ответ обоснуйте.`, + sol: `Графики линейных функций не имеют общих точек тогда и только тогда, когда прямые параллельны: угловые коэффициенты равны, свободные члены различны. +
Условие параллельности: $a - 3 = 2 \\implies a = 5$ +
Проверка: при $a = 5$ получаем $y = 2x + 2$ и $y = 2x - 2$. Угловые коэффициенты равны ($2 = 2$), свободные члены различны ($2 \\neq -2$) — прямые параллельны ✓ +
Ответ: $a = 5$
` + }, + { + text: `Диагонали ромба относятся как $3:4$, периметр ромба равен $200$ см. + Найдите площадь круга, окружность которого вписана в ромб.`, + sol: `Свойство диагоналей ромба: диагонали ромба взаимно перпендикулярны и точкой пересечения делятся пополам. +
Формула площади ромба: $S_{\\text{ромб}} = \\dfrac{d_{1}\\cdot d_{2}}{2}$. +
Формула радиуса вписанной окружности: $r = \\dfrac{S}{p}$ ($p$ — полупериметр). +
Формула площади круга: $S_{\\text{кр}} = \\pi r^{2}$. +
Шаг 1. По условию диагонали относятся как $3:4$. Введём параметр $k$: $d_{1} = 3k$, $d_{2} = 4k$. Половины диагоналей: $\\dfrac{3k}{2}$ и $2k$. Они являются катетами прямоугольного треугольника, гипотенуза которого — сторона ромба. По теореме Пифагора: +$$a = \\sqrt{\\left(\\dfrac{3k}{2}\\right)^{2} + (2k)^{2}} = \\sqrt{\\dfrac{9k^{2}}{4} + \\dfrac{16k^{2}}{4}} = \\sqrt{\\dfrac{25k^{2}}{4}} = \\dfrac{5k}{2}.$$ +Шаг 2. Из периметра $P = 4a = 200$ найдём сторону: $a = 50$ см. Тогда $\\dfrac{5k}{2} = 50 \\implies k = 20$. Значит: +$$d_{1} = 3\\cdot 20 = 60\\text{ см}, \\quad d_{2} = 4\\cdot 20 = 80\\text{ см}.$$ +Шаг 3. Площадь ромба: +$$S = \\dfrac{60\\cdot 80}{2} = 2400\\text{ см}^{2}.$$ +Шаг 4. Полупериметр $p = \\dfrac{P}{2} = 100$ см. Радиус вписанной окружности: +$$r = \\dfrac{S}{p} = \\dfrac{2400}{100} = 24\\text{ см}.$$ +Шаг 5. Площадь круга: +$$S_{\\text{кр}} = \\pi r^{2} = \\pi\\cdot 24^{2} = 576\\pi\\text{ см}^{2}.$$ + + + + + + + + + r=24 + 30 + 30 + 40 + 40 + +
Ответ: $576\\pi$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v76.js b/frontend/js/exam9/variants/v76.js new file mode 100644 index 0000000..644828a --- /dev/null +++ b/frontend/js/exam9/variants/v76.js @@ -0,0 +1,184 @@ +VARIANTS[76] = { + label: "Вариант 76", + tasks: [ + { + text: `Определите, какое из данных чисел является решением неравенства $x < -6$:`, + opts: [ + ["а", "$-3$"], ["б", "$-2$"], ["в", "$-1$"], ["г", "$0$"], ["д", "$-8$"], + ], + sol: `Неравенство $x \\lt -6$ выполняется только для чисел, строго меньших $-6$. +
    +
  • а) $-3$: $-3 \\gt -6$ — нет;
    • +
  • б) $-2$: нет;
    • +
  • в) $-1$: нет;
    • +
  • г) $0$: нет;
    • +
  • д) $-8$: $-8 \\lt -6$ — да ✓
    • +
+
Ответ: д) $-8$
` + }, + { + text: `Разность каких двух чисел НЕ равна $4{,}5$:`, + opts: [ + ["а", "$-0{,}5$ и $-5$"], ["б", "$-2{,}5$ и $-7$"], ["в", "$6{,}5$ и $2$"], + ["г", "$10$ и $5{,}5$"], ["д", "$-0{,}5$ и $4$"], + ], + sol: `Проверим разность в каждом варианте: +
    +
  • а) $-0{,}5 - (-5) = 4{,}5$ ✓
    • +
  • б) $-2{,}5 - (-7) = 4{,}5$ ✓
    • +
  • в) $6{,}5 - 2 = 4{,}5$ ✓
    • +
  • г) $10 - 5{,}5 = 4{,}5$ ✓
    • +
  • д) $-0{,}5 - 4 = -4{,}5 \\neq 4{,}5$  ✗ — НЕ равна
    • +
+
Ответ: д)
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "угол правильного шестиугольника равен $120^{\\circ}$;"], + ["б", "$\\operatorname{tg} 45^{\\circ} = \\operatorname{ctg} 45^{\\circ}$;"], + ["в", "центр окружности, вписанной в угол, равноудалён от сторон угла;"], + ["г", "в равностороннем треугольнике есть только два равных угла?"], + ], + sol: `
    +
  • а) Угол правильного шестиугольника $(6-2)\\cdot180^\\circ/6 = 120^\\circ$ — верно
    • +
  • б) $\\operatorname{tg}45^\\circ = \\operatorname{ctg}45^\\circ = 1$ — верно
    • +
  • в) Центр вписанной в угол окружности равноудалён от его сторон — верно
    • +
  • г) «В равностороннем треугольнике есть только два равных угла» — НЕВЕРНО: все три угла равны ($60^\\circ$ каждый).
    • +
+
Ответ: г)
` + }, + { + text: `Разложите на множители многочлен $64x^2 - 81y^2 + 8x - 9y$.`, + sol: `Первые два слагаемых — разность квадратов, последние два группируем: +$$64x^2 - 81y^2 + 8x - 9y = (64x^2 - 81y^2) + (8x - 9y)$$ +$$= (8x - 9y)(8x + 9y) + (8x - 9y)$$ +Выносим общий множитель $(8x - 9y)$: +$$= (8x - 9y)(8x + 9y + 1)$$ +
Ответ: $(8x - 9y)(8x + 9y + 1)$
` + }, + { + text: `Найдите значение выражения $a^2 : b - ab$ при $a = -4$, $b = -1{,}6$.`, + sol: `Порядок действий: сначала выполняем возведение в степень, затем умножение и деление, в конце сложение и вычитание. +
Свойство квадрата: $(-a)^2 = a^2$ (квадрат отрицательного числа положителен). +
Правило знаков: минус на минус даёт плюс. +
Шаг 1. Подставим $a = -4$ и $b = -1{,}6$ в выражение, заменив деление на дробь: +$$a^2 : b - ab = \\dfrac{a^2}{b} - ab.$$ +Шаг 2. Вычислим $a^2 = (-4)^2 = 16$ и подставим значения: +$$\\dfrac{16}{-1{,}6} - (-4)\\cdot(-1{,}6).$$ +Шаг 3. Выполним деление: +$$\\dfrac{16}{-1{,}6} = -10.$$ +Шаг 4. Выполним умножение (минус на минус — плюс): +$$(-4)\\cdot(-1{,}6) = 6{,}4.$$ +Шаг 5. Соберём всё (минус перед скобкой меняет знак): +$$-10 - 6{,}4 = -16{,}4.$$ +
Ответ: $-16{,}4$
` + }, + { + text: `Четвёртый член геометрической прогрессии равен $4$, а знаменатель равен $2$. + Найдите сумму четырёх первых членов этой прогрессии.`, + sol: `Формула $n$-го члена геометрической прогрессии: $a_n = a_1 \\cdot q^{n-1}$, откуда $a_1 = \\dfrac{a_n}{q^{n-1}}$. +
Связь соседних членов: $a_{n+1} = a_n \\cdot q$. +
Шаг 1. По условию $a_4 = 4$, $q = 2$. Найдём первый член из равенства $a_4 = a_1 \\cdot q^{3}$: +$$a_1 = \\dfrac{a_4}{q^{3}} = \\dfrac{4}{2^{3}} = \\dfrac{4}{8} = \\dfrac{1}{2}.$$ +Шаг 2. Найдём остальные члены, умножая каждый предыдущий на знаменатель $q = 2$: +$$a_2 = a_1\\cdot q = \\dfrac{1}{2}\\cdot 2 = 1, \\quad a_3 = a_2\\cdot q = 1\\cdot 2 = 2, \\quad a_4 = 4.$$ +Шаг 3. Сложим: +$$S_{4} = \\dfrac{1}{2} + 1 + 2 + 4 = \\dfrac{15}{2} = 7{,}5.$$ +
Ответ: $7{,}5$
` + }, + { + text: `В четырёхугольнике $ABCD$ $AB = 5$ см, $AD = 8$ см, $CD = 2\\sqrt{6}$ см, + $\\angle A = 60^{\\circ}$, $\\angle C = 90^{\\circ}$. Найдите длину стороны $BC$.`, + sol: `Теорема косинусов: $c^2 = a^2 + b^2 - 2ab\\cos\\gamma$. +
Теорема Пифагора: $c^2 = a^2 + b^2$ для прямоугольного треугольника. +
Значение: $\\cos 60^{\\circ} = \\dfrac{1}{2}$. +
Шаг 1. Проведём диагональ $BD$. Четырёхугольник разобьётся на два треугольника: $ABD$ (с известным углом $\\angle A = 60^{\\circ}$) и $BCD$ (с прямым углом $\\angle C = 90^{\\circ}$). +
Шаг 2. Найдём $BD$ из $\\triangle ABD$ по теореме косинусов. Известны $AB = 5$, $AD = 8$, $\\angle A = 60^{\\circ}$: +$$BD^2 = AB^2 + AD^2 - 2\\cdot AB\\cdot AD\\cdot\\cos A = 5^2 + 8^2 - 2\\cdot 5\\cdot 8\\cdot\\dfrac{1}{2}.$$ +Вычислим: $25 + 64 - 40 = 49$, значит $BD = \\sqrt{49} = 7$ см. +
Шаг 3. Найдём $BC$ из $\\triangle BCD$ по теореме Пифагора. Так как $\\angle C = 90^{\\circ}$, $BD$ — гипотенуза, а $BC$ и $CD$ — катеты. Подставим $CD = 2\\sqrt{6}$, $CD^2 = 4\\cdot 6 = 24$: +$$BC^2 = BD^2 - CD^2 = 49 - 24 = 25 \\implies BC = \\sqrt{25} = 5\\text{ см}.$$ + + + + + + A + D + B + C + 8 + 5 + 2√6 + 5 + BD=7 + 60° + +
Ответ: $BC = 5$ см
` + }, + { + text: `В первый день туристы прошли $0{,}3$ намеченного пути, + а во второй — преодолели $0{,}4$ намеченного пути. + В третий день им оставалось пройти последние $15$ км. + Каков весь путь туристов за $3$ дня?`, + sol: `Метод введения переменной: неизвестную величину обозначаем буквой и составляем уравнение. +
Шаг 1. Введём переменную. Пусть $S$ — весь намеченный путь (в километрах). +
Шаг 2. Выразим путь за два дня. В первый день туристы прошли $0{,}3\\cdot S$, во второй — $0{,}4\\cdot S$. Всего за два дня: +$$0{,}3S + 0{,}4S = 0{,}7S.$$ +Шаг 3. Выразим оставшийся путь. На третий день им осталось: +$$S - 0{,}7S = 0{,}3S.$$ +Шаг 4. Составим уравнение. По условию это $15$ км: +$$0{,}3S = 15.$$ +Шаг 5. Разделим обе части на $0{,}3$: +$$S = \\dfrac{15}{0{,}3} = 50\\text{ км}.$$ +
Ответ: $50$ км
` + }, + { + text: `При каком значении $a$ графики функций $y = (a+1)x + 6$ и $y = 3x - 2$ + не имеют общих точек? Ответ обоснуйте.`, + sol: `Условие параллельности прямых: две прямые $y = k_{1}x + b_{1}$ и $y = k_{2}x + b_{2}$ параллельны (и не имеют общих точек) тогда и только тогда, когда $k_{1} = k_{2}$ и $b_{1}\\neq b_{2}$. +
Шаг 1. Запишем угловые коэффициенты данных прямых: +$$k_{1} = a + 1\\quad\\text{(из первой формулы)},\\qquad k_{2} = 3\\quad\\text{(из второй)}.$$ +Шаг 2. Графики не имеют общих точек, когда прямые параллельны. Приравняем угловые коэффициенты: +$$a + 1 = 3 \\implies a = 2.$$ +Шаг 3. Проверим, что свободные члены не совпали (иначе прямые совпадают, а не параллельны). При $a = 2$: $y = 3x + 6$ и $y = 3x - 2$. Свободные члены $6$ и $-2$ различны — значит, прямые параллельны и не имеют общих точек ✓. +
Ответ: $a = 2$
` + }, + { + text: `Одна из диагоналей ромба в $1\\dfrac{1}{3}$ раза больше другой, + периметр ромба равен $100$ см. + Найдите площадь круга, окружность которого вписана в ромб.`, + sol: `Свойство диагоналей ромба: диагонали ромба взаимно перпендикулярны и делятся точкой пересечения пополам. +
Формула площади ромба: $S = \\dfrac{d_{1}\\cdot d_{2}}{2}$. +
Формула радиуса вписанной окружности: $r = \\dfrac{S}{p}$ ($p$ — полупериметр). +
Формула площади круга: $S_{\\text{кр}} = \\pi r^{2}$. +
Шаг 1. Заметим, что $1\\dfrac{1}{3} = \\dfrac{4}{3}$. Пусть меньшая диагональ $d_{1}$, тогда большая $d_{2} = \\dfrac{4}{3}d_{1}$. Половины диагоналей — катеты прямоугольного треугольника, гипотенуза которого — сторона ромба: +$$\\dfrac{d_{1}}{2} \\quad\\text{и}\\quad \\dfrac{d_{2}}{2} = \\dfrac{4d_{1}}{6} = \\dfrac{2d_{1}}{3}.$$ +По теореме Пифагора: +$$a = \\sqrt{\\left(\\dfrac{d_{1}}{2}\\right)^{2} + \\left(\\dfrac{2d_{1}}{3}\\right)^{2}} = d_{1}\\sqrt{\\dfrac{1}{4} + \\dfrac{4}{9}} = d_{1}\\sqrt{\\dfrac{9 + 16}{36}} = d_{1}\\cdot\\dfrac{5}{6}.$$ +Шаг 2. Из периметра $P = 4a = 100$ имеем $a = 25$ см. Тогда $d_{1}\\cdot\\dfrac{5}{6} = 25 \\implies d_{1} = 30$ см, $d_{2} = \\dfrac{4}{3}\\cdot 30 = 40$ см. +
Шаг 3. Площадь ромба: +$$S = \\dfrac{30\\cdot 40}{2} = 600\\text{ см}^{2}.$$ +Шаг 4. Полупериметр $p = \\dfrac{P}{2} = 50$ см. Радиус вписанной окружности: +$$r = \\dfrac{S}{p} = \\dfrac{600}{50} = 12\\text{ см}.$$ +Шаг 5. Площадь круга: +$$S_{\\text{кр}} = \\pi r^{2} = \\pi\\cdot 12^{2} = 144\\pi\\text{ см}^{2}.$$ + + + + + + + + + r=12 + 15 + 15 + 20 + 20 + +
Ответ: $144\\pi$ см²
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v77.js b/frontend/js/exam9/variants/v77.js new file mode 100644 index 0000000..c0ccdd4 --- /dev/null +++ b/frontend/js/exam9/variants/v77.js @@ -0,0 +1,220 @@ +VARIANTS[77] = { + label: "Вариант 77", + tasks: [ + { + text: `Определите, решением какого из данных неравенств является числовой промежуток $(-\\infty;\\; -3]$:`, + opts: [ + ["а", "$x < -3$"], ["б", "$x > -3$"], ["в", "$x \\leq -3$"], + ["г", "$x \\geq -3$"], ["д", "$x \\leq 3$"], + ], + sol: `Промежуток $(-\\infty;\\,-3]$ включает все числа, не превосходящие $-3$, то есть $x\\leq -3$. +
    +
  • а) $x \\lt -3$ — не включает $-3$ (открытый);
    • +
  • б) $x \\gt -3$ — правее $-3$;
    • +
  • в) $x\\leq -3$ — совпадает с $(-\\infty;\\,-3]$ ✓
    • +
  • г) $x\\geq -3$ — правее или равно $-3$;
    • +
  • д) $x\\leq 3$ — совсем другой промежуток.
    • +
+
Ответ: в)
` + }, + { + text: `Произведение каких двух чисел НЕ равно $-5$:`, + opts: [ + ["а", "$1$ и $-5$"], ["б", "$-2$ и $2{,}5$"], ["в", "$-0{,}5$ и $10$"], + ["г", "$1$ и $5$"], ["д", "$-1$ и $5$"], + ], + sol: `Проверяем каждую пару: +
    +
  • а) $1\\cdot(-5)=-5$ ✓
    • +
  • б) $(-2)\\cdot2{,}5=-5$ ✓
    • +
  • в) $(-0{,}5)\\cdot10=-5$ ✓
    • +
  • г) $1\\cdot5=5\\neq-5$ — НЕ равно $-5$
    • +
  • д) $(-1)\\cdot5=-5$ ✓
    • +
+
Ответ: г)
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "в равнобедренном треугольнике два угла равны;"], + ["б", "площадь равностороннего треугольника со стороной $a$ равна $S = \\dfrac{a^2\\sqrt{3}}{4}$;"], + ["в", "около любого четырёхугольника можно описать окружность;"], + ["г", "вертикальные углы равны между собой?"], + ], + sol: `
    +
  • а) В равнобедренном треугольнике углы при основании равны — верно
    • +
  • б) $S=\\dfrac{a^2\\sqrt{3}}{4}$ для равностороннего треугольника — верно
    • +
  • в) Около любого четырёхугольника можно описать окружность — НЕВЕРНО
    + Описанная окружность существует только у вписанных (циклических) четырёхугольников, у которых сумма противоположных углов равна $180^\\circ$. Например, около произвольного параллелограмма (не являющегося прямоугольником) описать окружность нельзя.
    • +
  • г) Вертикальные углы равны — верно
    • +
+
Ответ: в)
` + }, + { + text: `Определите масштаб изображения, если расстояние на местности, + равное $25$ км, изображено на карте отрезком в $2{,}5$ мм.`, + sol: `Переведём расстояние на местности в миллиметры: +$$25\\text{ км} = 25\\cdot1\\,000\\,000\\text{ мм} = 25\\,000\\,000\\text{ мм}$$ +Масштаб — отношение длины на карте к длине на местности: +$$M = \\dfrac{2{,}5\\text{ мм}}{25\\,000\\,000\\text{ мм}} = \\dfrac{1}{10\\,000\\,000}$$ +
Ответ: $1:10\\,000\\,000$
` + }, + { + text: `Сумма градусных мер вписанного угла и дуги, на которую он опирается, равна $120^{\\circ}$. + Найдите градусную меру вписанного угла.`, + sol: `Теорема о вписанном угле: вписанный угол равен половине дуги, на которую он опирается. Значит, дуга равна удвоенному вписанному углу. +
Шаг 1. Обозначим вписанный угол через $\\alpha$ (в градусах). Тогда соответствующая дуга равна $2\\alpha$. +
Шаг 2. По условию сумма градусной меры вписанного угла и дуги равна $120^\\circ$: +$$\\alpha + 2\\alpha = 120^\\circ.$$ +Шаг 3. Приведём подобные и найдём $\\alpha$: +$$3\\alpha = 120^\\circ \\implies \\alpha = \\dfrac{120^\\circ}{3} = 40^\\circ.$$ +
Ответ: $40^\\circ$
` + }, + { + text: `Найдите координаты точки графика линейной функции $y = 2x - 35$, + абсцисса которой в $3$ раза больше ординаты.`, + sol: `Метод подстановки: если переменные связаны равенством, одну из них выражаем через другую и подставляем в уравнение. +
Координаты точки графика: абсцисса — это $x$, ордината — это $y$. +
Шаг 1. Запишем условие задачи в виде равенства. «Абсцисса в $3$ раза больше ординаты» означает, что $x = 3y$. +
Шаг 2. Подставим $x = 3y$ в уравнение функции $y = 2x - 35$: +$$y = 2\\cdot 3y - 35 = 6y - 35.$$ +Шаг 3. Перенесём $6y$ в левую часть, поменяв знак: +$$y - 6y = -35 \\implies -5y = -35.$$ +Шаг 4. Разделим обе части на $-5$: +$$y = \\dfrac{-35}{-5} = 7.$$ +Шаг 5. Найдём абсциссу из $x = 3y$: +$$x = 3\\cdot 7 = 21.$$ +Проверка. Подставим $x = 21$ в формулу функции: $y = 2\\cdot 21 - 35 = 42 - 35 = 7$ ✓. Условие $21 = 3\\cdot 7$ ✓. +
Ответ: $(21;\\;7)$
` + }, + { + text: `Известно, что функция $y = f(x)$ является чётной и $f(3) = -7$, $f(-4) = 5$. + Найдите значение выражения $2f(-3) + 3f(4)$.`, + sol: `Свойство чётной функции: $f(-x) = f(x)$ для всех $x$ из области определения (график симметричен относительно оси $Oy$). +
Шаг 1. Найдём $f(-3)$. По свойству чётности значение в точке $-3$ равно значению в точке $3$: +$$f(-3) = f(3) = -7.$$ +Шаг 2. Найдём $f(4)$. Аналогично, $f(4) = f(-4) = 5$ (значения в противоположных точках равны). +
Шаг 3. Подставим найденные значения в выражение $2f(-3) + 3f(4)$: +$$2f(-3) + 3f(4) = 2\\cdot(-7) + 3\\cdot 5 = -14 + 15 = 1.$$ +
Ответ: $1$
` + }, + { + text: `Решите уравнение $\\dfrac{x^2 - 5x}{x - 5} = 2 - x^2$.`, + sol: `Правило сокращения дроби: если в числителе и знаменателе есть одинаковый множитель, его можно сократить (при условии, что он не равен нулю). +
Формула корней квадратного уравнения: $x = \\dfrac{-b \\pm \\sqrt{D}}{2a}$, где $D = b^2 - 4ac$ — дискриминант. +
Шаг 1. ОДЗ. Знаменатель $x - 5 \\neq 0$, значит, $x \\neq 5$. +
Шаг 2. Вынесем в числителе общий множитель $x$ и сократим дробь: +$$\\dfrac{x^2 - 5x}{x - 5} = \\dfrac{x(x - 5)}{x - 5} = x.$$ +Шаг 3. Уравнение принимает вид $x = 2 - x^2$. Перенесём всё в одну сторону: +$$x^2 + x - 2 = 0.$$ +Шаг 4. Найдём дискриминант ($a = 1$, $b = 1$, $c = -2$): +$$D = 1^{2} - 4\\cdot 1\\cdot(-2) = 1 + 8 = 9.$$ +Шаг 5. Найдём корни: +$$x = \\dfrac{-1 \\pm \\sqrt{9}}{2} = \\dfrac{-1 \\pm 3}{2}.$$ +Получаем $x_{1} = \\dfrac{-1+3}{2} = 1$ и $x_{2} = \\dfrac{-1-3}{2} = -2$. +
Шаг 6. Проверим по ОДЗ: оба корня не равны $5$, значит, оба подходят. +
Ответ: $x = 1,\\quad x = -2$
` + }, + { + text: `В треугольнике $ABC$ проведены биссектриса $BK$ и медиана $AM$, + которые пересекаются в точке $F$. + Площадь треугольника $ABC$ равна $210$, $AB : BC = 3 : 4$. + Найдите площадь четырёхугольника $KFMC$.`, + sol: ` + + + + + + + + + + A + B + C + + + M + + + K + + + F + + KFMC + +Шаг 1. По теореме о биссектрисе: $AK:KC = AB:BC = 3:4$. +Из вершины $B$ проведём высоту к $AC$ — основание общее для $\\triangle ABK$ и $\\triangle CBK$: +$$S(ABK) = \\dfrac{AK}{AC}\\cdot S(ABC) = \\dfrac{3}{7}\\cdot210 = 90,\\quad S(CBK) = 120$$ +Шаг 2. $AM$ — медиана, $M$ — середина $BC$: +$$S(ABM) = S(ACM) = \\dfrac{210}{2} = 105$$ +Шаг 3. Находим $AF:FM$ через параллельную. +
Проведём через $M$ прямую $MN \\parallel BK$, где $N$ — на $AC$. + + + + + + + + + + + A + B + C + M + K + F + N + KFMC + AK=3/7 + KN=2/7 + +По теореме о средней линии в $\\triangle BCK$ ($M$ — середина $BC$, $MN \\parallel BK$): +
$N$ — середина $CK$, т.е. $KN = \\dfrac{1}{2}CK = \\dfrac{1}{2}\\cdot\\dfrac{4}{7}AC = \\dfrac{2}{7}AC$. +
Таким образом: $AK = \\dfrac{3}{7}AC$, $KN = \\dfrac{2}{7}AC$. +
По теореме Фалеса (две параллельные $BK$ и $MN$ пересекают две секущие $AM$ и $AC$ из точки $A$): +$$\\dfrac{AF}{FM} = \\dfrac{AK}{KN} = \\dfrac{3/7}{2/7} = \\dfrac{3}{2} \\implies AF:FM = 3:2$$ +
Шаг 4. +$$S(ABF) = \\dfrac{AF}{AM}\\cdot S(ABM) = \\dfrac{3}{5}\\cdot105 = 63$$ +$$S(BFM) = S(ABM) - S(ABF) = 105 - 63 = 42$$ +Шаг 5. +$$S(ACF) = \\dfrac{AF}{AM}\\cdot S(ACM) = \\dfrac{3}{5}\\cdot105 = 63$$ +$$S(AKF) = \\dfrac{AK}{AC}\\cdot S(ACF) = \\dfrac{3}{7}\\cdot63 = 27$$ +Шаг 6. +$$S(KFMC) = S(ABC) - S(ABF) - S(BFM) - S(AKF) = 210 - 63 - 42 - 27 = 78$$ +
Ответ: $78$
` + }, + { + text: `В зрительном зале было $320$ мест, причём в каждом ряду их было одинаковое количество. + Число рядов уменьшили на $2$, а в каждый ряд добавили $5$ мест. + В результате в зале стало $350$ мест. + Сколько рядов стало в зрительном зале?`, + sol: `Метод введения двух переменных: вводим переменные для каждой неизвестной величины и составляем систему уравнений. +
Формула корней квадратного уравнения: $x = \\dfrac{-b\\pm\\sqrt{D}}{2a}$, где $D = b^2 - 4ac$. +
Шаг 1. Введём переменные. Пусть $r$ — первоначальное число рядов, $n$ — количество мест в одном ряду. Тогда всего мест: +$$r\\cdot n = 320.\\qquad(1)$$ +Шаг 2. Составим второе уравнение. После изменений число рядов стало $(r-2)$, мест в ряду — $(n+5)$, всего $350$ мест: +$$(r - 2)(n + 5) = 350.$$ +Шаг 3. Раскроем скобки: +$$rn + 5r - 2n - 10 = 350.$$ +Подставим $rn = 320$ из (1): +$$320 + 5r - 2n - 10 = 350 \\implies 5r - 2n = 40.\\qquad(2)$$ +Шаг 4. Решим систему. Из (1) выразим $n = \\dfrac{320}{r}$ и подставим в (2): +$$5r - 2\\cdot\\dfrac{320}{r} = 40 \\implies 5r - \\dfrac{640}{r} = 40.$$ +Шаг 5. Умножим обе части на $r$ (заметим, что $r\\neq 0$, так как число рядов положительное): +$$5r^2 - 640 = 40r \\implies 5r^2 - 40r - 640 = 0.$$ +Разделим на $5$: +$$r^2 - 8r - 128 = 0.$$ +Шаг 6. Найдём дискриминант ($a=1$, $b=-8$, $c=-128$): +$$D = (-8)^2 - 4\\cdot 1\\cdot(-128) = 64 + 512 = 576 = 24^2.$$ +$$r = \\dfrac{8\\pm 24}{2}: \\quad r_{1} = \\dfrac{32}{2} = 16,\\quad r_{2} = \\dfrac{-16}{2} = -8.$$ +По смыслу задачи $r \\gt 0$, поэтому $r = 16$. +
Шаг 7. В вопросе спрашивается, сколько рядов стало: это $r - 2 = 16 - 2 = 14$. +
Ответ: $14$ рядов
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v78.js b/frontend/js/exam9/variants/v78.js new file mode 100644 index 0000000..c3ec68f --- /dev/null +++ b/frontend/js/exam9/variants/v78.js @@ -0,0 +1,163 @@ +VARIANTS[78] = { + label: "Вариант 78", + tasks: [ + { + text: `Определите, решением какого из данных неравенств является числовой промежуток $(-4;\\; +\\infty)$:`, + opts: [ + ["а", "$x < -4$"], ["б", "$x > -4$"], ["в", "$x \\leq -4$"], + ["г", "$x \\geq -4$"], ["д", "$x \\leq 4$"], + ], + sol: `Промежуток $(-4;\\,+\\infty)$ — все числа строго больше $-4$. Неравенство $x \\gt -4$ задаёт его. +
Ответ: б)
` + }, + { + text: `Частное каких двух чисел НЕ равно $-2$:`, + opts: [ + ["а", "$10$ и $-5$"], ["б", "$-2$ и $1$"], ["в", "$-1$ и $0{,}5$"], + ["г", "$-0{,}5$ и $1$"], ["д", "$-6$ и $3$"], + ], + sol: `а)$-2$✓ б)$-2$✓ в)$-2$✓ г)$-0{,}5\\neq-2$ НЕ равно д)$-2$✓ +
Ответ: г)
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "если в треугольнике два угла равны, то он — равнобедренный;"], + ["б", "площадь прямоугольного треугольника с катетами $a$ и $b$ равна $S = \\dfrac{ab}{2}$;"], + ["в", "в любой четырёхугольник можно вписать окружность;"], + ["г", "сумма смежных углов равна $180^{\\circ}$?"], + ], + sol: `а) верно; б) верно; г) верно. +
в) НЕВЕРНО — вписанная окружность есть только если $AB+CD=BC+AD$. +
Ответ: в)
` + }, + { + text: `Определите масштаб изображения, если расстояние на местности, + равное $50$ км, изображено на карте отрезком в $5$ мм.`, + sol: `Переводим: $50$ км $= 50\\,000\\,000$ мм. +$$M = \\dfrac{5\\text{ мм}}{50\\,000\\,000\\text{ мм}} = \\dfrac{1}{10\\,000\\,000}$$ +
Ответ: $1:10\\,000\\,000$
` + }, + { + text: `Сумма градусных мер вписанного угла и соответствующего ему центрального угла равна $150^{\\circ}$. + Найдите градусную меру вписанного угла.`, + sol: `Теорема о вписанном угле: вписанный угол равен половине центрального угла, опирающегося на ту же дугу. Значит, центральный угол вдвое больше вписанного. +
Шаг 1. Обозначим вписанный угол через $\\alpha$. Тогда соответствующий центральный угол равен $2\\alpha$. +
Шаг 2. По условию их сумма равна $150^\\circ$: +$$\\alpha + 2\\alpha = 150^\\circ.$$ +Шаг 3. Приведём подобные и найдём $\\alpha$: +$$3\\alpha = 150^\\circ \\implies \\alpha = 50^\\circ.$$ +
Ответ: $50^\\circ$
` + }, + { + text: `Найдите координаты точки графика линейной функции $y = -4x - 35$, + ордината которой в $3$ раза больше абсциссы.`, + sol: `Метод подстановки: используя связь между координатами, выражаем одну переменную через другую и подставляем в уравнение функции. +
Координаты точки: абсцисса — это $x$, ордината — это $y$. +
Шаг 1. Запишем условие в виде равенства. «Ордината в $3$ раза больше абсциссы» означает $y = 3x$. +
Шаг 2. Подставим $y = 3x$ в уравнение функции $y = -4x - 35$: +$$3x = -4x - 35.$$ +Шаг 3. Перенесём $-4x$ влево с противоположным знаком: +$$3x + 4x = -35 \\implies 7x = -35.$$ +Шаг 4. Разделим обе части на $7$: +$$x = \\dfrac{-35}{7} = -5.$$ +Шаг 5. Найдём ординату из $y = 3x$: +$$y = 3\\cdot(-5) = -15.$$ +Проверка. Подставим $x = -5$: $y = -4\\cdot(-5) - 35 = 20 - 35 = -15$ ✓. +
Ответ: $(-5;\\;-15)$
` + }, + { + text: `Известно, что функция $y = f(x)$ является нечётной и $f(5) = -2$, $f(-7) = 8$. + Найдите значение выражения $2f(-5) + 3f(7)$.`, + sol: `Свойство нечётной функции: $f(-x) = -f(x)$ для всех $x$ из области определения (график симметричен относительно начала координат). +
Шаг 1. Найдём $f(-5)$. По свойству нечётности: +$$f(-5) = -f(5) = -(-2) = 2.$$ +Шаг 2. Найдём $f(7)$. По свойству нечётности: +$$f(7) = -f(-7) = -8.$$ +Шаг 3. Подставим в выражение $2f(-5) + 3f(7)$: +$$2\\cdot 2 + 3\\cdot(-8) = 4 + (-24) = -20.$$ +
Ответ: $-20$
` + }, + { + text: `Решите уравнение $\\dfrac{x^2 + 3x}{x + 3} = 2 - x^2$.`, + sol: `Правило сокращения дроби: одинаковый множитель в числителе и знаменателе сокращается (если он не равен нулю). +
Теорема Виета: для уравнения $x^2 + px + q = 0$ сумма корней равна $-p$, произведение равно $q$. +
Шаг 1. ОДЗ. Знаменатель $x + 3 \\neq 0$, значит, $x \\neq -3$. +
Шаг 2. Вынесем в числителе общий множитель $x$ и сократим: +$$\\dfrac{x^2 + 3x}{x + 3} = \\dfrac{x(x + 3)}{x + 3} = x.$$ +Шаг 3. Уравнение становится $x = 2 - x^2$, или +$$x^2 + x - 2 = 0.$$ +Шаг 4. По теореме Виета подбираем два числа: сумма $-1$, произведение $-2$ — это $1$ и $-2$. Значит, $x_{1} = 1$, $x_{2} = -2$. +
Шаг 5. Проверим по ОДЗ: ни один корень не равен $-3$ — оба подходят. +
Ответ: $x=1,\\ x=-2$
` + }, + { + text: `В треугольнике $ABC$ проведены биссектриса $BK$ и медиана $CM$, + которые пересекаются в точке $F$. + Площадь треугольника $ABC$ равна $120$, $AB : BC = 2 : 3$. + Найдите площадь четырёхугольника $AMFK$.`, + sol: ` + + + + + A + B + C + + M + + K + + F + AMFK + +Свойство биссектрисы: биссектриса делит противоположную сторону в отношении прилежащих сторон. Здесь $BK$ — биссектриса из $B$: $AK:KC = AB:BC = 2:3$. +
Свойство медианы: медиана $CM$ из $C$ делит сторону $AB$ пополам — $M$ — середина $AB$. +
Свойство площадей: если у треугольников общая высота, то их площади относятся как основания. +
Теорема Фалеса: параллельные прямые отсекают на двух пересекающих их прямых пропорциональные отрезки. +
Шаг 1. $AK : KC = 2 : 3$, поэтому $AK = \\dfrac{2}{5}AC$. Треугольники $ABK$ и $ABC$ имеют общую высоту из $B$, значит площади относятся как основания: +$$S(ABK) = \\dfrac{AK}{AC}\\cdot S(ABC) = \\dfrac{2}{5}\\cdot 120 = 48.$$ +Шаг 2. Медиана $CM$ делит $\\triangle ABC$ на два равновеликих: +$$S(ACM) = \\dfrac{1}{2}\\cdot S(ABC) = 60.$$ +Шаг 3. Найдём отношение $CF : FM$. Через $M$ проведём прямую $MN \\parallel BK$, $N$ — на $AC$. По теореме Фалеса (в $\\triangle ABK$ прямая через $M$, параллельная $BK$, отсекает середину $AK$): $N$ — середина $AK$, поэтому $AN = NK = \\dfrac{1}{2}AK = \\dfrac{1}{5}AC$. +
В $\\triangle ACM$ прямые $BK$ и $MN$ параллельны, их секут $CM$ (в точках $F$ и $M$) и $CA$ (в точках $K$ и $N$). По теореме Фалеса: +$$\\dfrac{CF}{FM} = \\dfrac{CK}{KN} = \\dfrac{3/5\\cdot AC}{1/5\\cdot AC} = \\dfrac{3}{1}.$$ +Шаг 4. В $\\triangle ACM$ точка $F$ делит $CM$ так, что $CF = \\dfrac{3}{4}CM$, $FM = \\dfrac{1}{4}CM$: +$$S(ACF) = \\dfrac{CF}{CM}\\cdot S(ACM) = \\dfrac{3}{4}\\cdot 60 = 45,$$ +$$S(AMF) = S(ACM) - S(ACF) = 60 - 45 = 15.$$ +Шаг 5. В $\\triangle ACF$ точка $K$ на $AC$ делит его в отношении $AK : KC = 2 : 3$: +$$S(AKF) = \\dfrac{AK}{AC}\\cdot S(ACF) = \\dfrac{2}{5}\\cdot 45 = 18.$$ +Шаг 6. Четырёхугольник $AMFK$ состоит из двух треугольников $AMF$ и $AKF$ с общей стороной $AF$: +$$S(AMFK) = S(AMF) + S(AKF) = 15 + 18 = 33.$$ +
Ответ: $33$
` + }, + { + text: `Имеется $2$ ящика для упаковки фруктов. В первом ящике — $54$ ячейки, + причём в каждом ряду их одинаковое количество. Во втором ящике — $56$ ячеек, + при этом число рядов меньше на $2$, а в каждом ряду на $5$ ячеек больше, чем в первом. + Сколько ячеек в каждом ряду в первом ящике?`, + sol: `Метод введения двух переменных: вводим переменные для неизвестных и составляем систему уравнений. +
Формула корней квадратного уравнения: $x = \\dfrac{-b\\pm\\sqrt{D}}{2a}$, $D = b^2 - 4ac$. +
Шаг 1. Введём переменные для первого ящика. Пусть $r$ — число рядов, $n$ — число ячеек в одном ряду. Тогда всего ячеек: +$$r\\cdot n = 54.\\qquad(1)$$ +Шаг 2. Составим уравнение для второго ящика. Число рядов меньше на $2$, в каждом ряду на $5$ ячеек больше, всего $56$ ячеек: +$$(r-2)(n+5) = 56.$$ +Шаг 3. Раскроем скобки: +$$rn + 5r - 2n - 10 = 56.$$ +Подставим $rn = 54$ из (1): +$$54 + 5r - 2n - 10 = 56 \\implies 5r - 2n = 12.\\qquad(2)$$ +Шаг 4. Решим систему. Из (1) выразим $n = \\dfrac{54}{r}$ и подставим в (2): +$$5r - \\dfrac{108}{r} = 12.$$ +Шаг 5. Умножим на $r$ (при $r\\neq 0$): +$$5r^2 - 108 = 12r \\implies 5r^2 - 12r - 108 = 0.$$ +Шаг 6. Найдём дискриминант ($a = 5$, $b = -12$, $c = -108$): +$$D = (-12)^2 - 4\\cdot 5\\cdot(-108) = 144 + 2160 = 2304 = 48^2.$$ +$$r = \\dfrac{12\\pm 48}{10}:\\quad r_{1} = \\dfrac{60}{10} = 6,\\quad r_{2} = \\dfrac{-36}{10} \\lt 0.$$ +По смыслу задачи $r \\gt 0$, поэтому $r = 6$. +
Шаг 7. Найдём $n$ из (1): $n = \\dfrac{54}{6} = 9$ ячеек в ряду. +
Проверка для второго ящика: $(6-2)(9+5) = 4\\cdot 14 = 56$ ✓. +
Ответ: $9$ ячеек в ряду
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v79.js b/frontend/js/exam9/variants/v79.js new file mode 100644 index 0000000..8bf8a44 --- /dev/null +++ b/frontend/js/exam9/variants/v79.js @@ -0,0 +1,181 @@ +VARIANTS[79] = { + label: "Вариант 79", + tasks: [ + { + text: `Определите наибольшее из значений числовых выражений:`, + opts: [ + ["а", "$2^2$"], ["б", "$3^{-1}$"], ["в", "$(-2)^{-1}$"], + ["г", "$(-4)^2$"], ["д", "$\\left(\\dfrac{1}{8}\\right)^{-1}$"], + ], + sol: `Вычислим каждое: а) $2^2=4$; б) $3^{-1}=\\dfrac{1}{3}$; в) $(-2)^{-1}=-\\dfrac{1}{2}$; г) $(-4)^2=16$; д) $\\left(\\dfrac{1}{8}\\right)^{-1}=8$. +
Наибольшее: $16$. +
Ответ: г) $(-4)^2 = 16$
` + }, + { + text: `Наименьшим целым решением неравенства $2x > -5$ является:`, + opts: [ + ["а", "$-3$"], ["б", "$-2{,}5$"], ["в", "$-2$"], ["г", "$-1$"], ["д", "$0$"], + ], + sol: `$2x > -5 \\implies x > -2{,}5$. +
Наименьшее целое, строго большее $-2{,}5$: $x=-2$. +
Ответ: в) $-2$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "площадь треугольника можно найти по формуле $S = \\dfrac{1}{2} a h_a$;"], + ["б", "диагонали равнобедренной трапеции равны между собой;"], + ["в", "$\\operatorname{ctg} 45^{\\circ} = 1$;"], + ["г", "окружность, описанная около четырёхугольника, касается всех его сторон?"], + ], + sol: `
    +
  • а) $S=\\tfrac{1}{2}ah_a$ — верно
    • +
  • б) Диагонали равнобедренной трапеции равны — верно
    • +
  • в) $\\operatorname{ctg}45°=1$ — верно
    • +
  • г) «Описанная окружность касается всех сторон» — НЕВЕРНО. Описанная окружность проходит через вершины (а касается сторон вписанная).
    • +
+
Ответ: г)
` + }, + { + text: `Упростите выражение $\\dfrac{x^2-a^2}{x+a} - (3x-a)$ + и найдите его значение при $x = 12$.`, + sol: `Раскладываем числитель по формуле разности квадратов ($x\\neq -a$): +$$\\dfrac{x^2-a^2}{x+a} - (3x-a) = \\dfrac{(x-a)(x+a)}{x+a} - (3x-a) = (x-a)-(3x-a)$$ +$$= x - a - 3x + a = -2x$$ +При $x=12$: $-2\\cdot12 = -24$. +
Ответ: $-2x$;  при $x=12$ значение равно $-24$
` + }, + { + text: `При каком значении $x$ числа $x+1$, $2x-3$, $6x+6$ + являются последовательными членами арифметической прогрессии?`, + sol: `Характеристическое свойство арифметической прогрессии: каждый член (начиная со второго) равен среднему арифметическому соседних. Если $a$, $b$, $c$ — три подряд идущих члена АП, то $b = \\dfrac{a+c}{2}$, или равносильно $2b = a + c$. +
Шаг 1. Применим свойство к трём данным числам $a = x+1$, $b = 2x-3$, $c = 6x+6$: +$$2(2x - 3) = (x + 1) + (6x + 6).$$ +Шаг 2. Раскроем скобки и приведём подобные: +$$4x - 6 = 7x + 7.$$ +Шаг 3. Перенесём $x$-ы в одну сторону, числа в другую: +$$4x - 7x = 7 + 6,$$ +$$-3x = 13.$$ +Шаг 4. Разделим на $-3$: +$$x = -\\dfrac{13}{3}.$$ +Проверка. При $x = -\\dfrac{13}{3}$ члены прогрессии: $x+1 = -\\dfrac{10}{3}$, $2x-3 = -\\dfrac{35}{3}$, $6x+6 = -\\dfrac{60}{3} = -20$. Разности: $-\\dfrac{35}{3} - \\left(-\\dfrac{10}{3}\\right) = -\\dfrac{25}{3}$ и $-\\dfrac{60}{3} - \\left(-\\dfrac{35}{3}\\right) = -\\dfrac{25}{3}$ — разности равны, прогрессия ✓. +
Ответ: $x = -\\dfrac{13}{3}$
` + }, + { + text: `Найдите площадь треугольника $ABC$, если размеры одной клетки $1$ см $\\times$ $1$ см.`, + figure: ``, + sol: `Формула площади треугольника по координатам вершин (формула «шнурков»): +$$S = \\dfrac{1}{2}\\bigl|x_{A}(y_{B}-y_{C}) + x_{B}(y_{C}-y_{A}) + x_{C}(y_{A}-y_{B})\\bigr|.$$ +Альтернатива — метод «описанного прямоугольника»: описать вокруг треугольника прямоугольник со сторонами по линиям сетки, его площадь подсчитать по клеткам, а затем вычесть площади трёх прямоугольных треугольников по углам. +
Шаг 1. По рисунку определить координаты вершин $A$, $B$, $C$ в клетках. +
Шаг 2. Подставить координаты в формулу или применить метод описанного прямоугольника. +
Шаг 3. Получить ответ в см² (так как одна клетка — $1$ см $\\times$ $1$ см, площадь клетки $= 1$ см²). +
Ответ: определяется по рисунку
` + }, + { + text: `Из всех учащихся, выполнявших контрольную работу, на «$10$» выполнили $2$ учащихся, + на «$9$» — $3$, на «$8$» — $5$, на «$7$» — $6$, на «$6$» — $3$, на «$5$» — $1$. + Какой процент всех учащихся составляют учащиеся, получившие оценки не меньше «$7$»?`, + sol: `Формула вычисления процентного отношения: чтобы найти, какой процент составляет число $a$ от числа $b$, надо отношение $\\dfrac{a}{b}$ умножить на $100\\%$. +
Шаг 1. Найдём общее число учащихся, сложив количество получивших каждую оценку: +$$N = 2 + 3 + 5 + 6 + 3 + 1 = 20.$$ +Шаг 2. Найдём число учащихся с оценками не меньше «$7$». Это оценки $7$, $8$, $9$, $10$ — соответственно $6$, $5$, $3$, $2$ учащихся: +$$N_{\\geq 7} = 6 + 5 + 3 + 2 = 16.$$ +Шаг 3. Найдём процентное отношение: +$$\\dfrac{N_{\\geq 7}}{N}\\cdot 100\\% = \\dfrac{16}{20}\\cdot 100\\% = 0{,}8\\cdot 100\\% = 80\\%.$$ +
Ответ: $80\\%$
` + }, + { + text: `Не выполняя построения, найдите координаты точек пересечения параболы + $y = 2x^2 + 3$ и прямой $y = 2x + 7$.`, + sol: `Метод поиска точек пересечения графиков: в точках пересечения значения функций совпадают, поэтому приравниваем правые части и получаем уравнение относительно $x$. +
Теорема Виета (обратная): если $x_{1}+x_{2}=-p$ и $x_{1}\\cdot x_{2}=q$, то $x_{1}$ и $x_{2}$ — корни уравнения $x^{2}+px+q=0$. +
Шаг 1. Приравняем правые части уравнений: +$$2x^2 + 3 = 2x + 7.$$ +Шаг 2. Перенесём всё в левую часть: +$$2x^2 - 2x - 4 = 0.$$ +Разделим обе части на $2$: +$$x^2 - x - 2 = 0.$$ +Шаг 3. По теореме Виета ищем два числа, у которых сумма $1$, а произведение $-2$. Подходят $2$ и $-1$. Значит, $(x-2)(x+1) = 0$, откуда $x_{1} = 2$, $x_{2} = -1$. +
Шаг 4. Найдём ординаты точек пересечения, подставив корни в более простое уравнение прямой $y = 2x + 7$: +$$\\text{при } x = 2:\\;\\; y = 2\\cdot 2 + 7 = 11;$$ +$$\\text{при } x = -1:\\;\\; y = 2\\cdot(-1) + 7 = 5.$$ +
Ответ: $(2;\\,11)$ и $(-1;\\,5)$
` + }, + { + text: `В прямоугольнике $ABCD$ $AB = 3$ см, $AD = 4$ см. В треугольники $ABC$ и $ADC$ + вписаны окружности, которые касаются диагонали $AC$ в точках $M$ и $K$. + Найдите длину отрезка $MK$.`, + figure: ``, + sol: `Диагональ $AC = \\sqrt{AB^2+BC^2} = \\sqrt{9+16} = 5$ см. +
Оба треугольника $ABC$ и $ACD$ прямоугольные с катетами $3$ и $4$, гипотенузой $5$. + + + + + + + + + + + + + + + + + + + + + + + A + B + C + D + + M + K + + 2 + 1 + 2 + + AB = 3 + BC = 4 + + AC=5 + +Шаг 1. Радиус вписанной окружности прямоугольного треугольника: +$$r = \\dfrac{a+b-c}{2} = \\dfrac{3+4-5}{2} = 1\\text{ см}$$ +Шаг 2. Точка касания $M$ на $AC$ в $\\triangle ABC$. +
Касательная из $A$: $s - BC = \\dfrac{3+4+5}{2} - 4 = 2$. Значит $AM = 2$ см. +
Шаг 3. Точка касания $K$ на $AC$ в $\\triangle ACD$. +
Касательная из $A$: $s - CD = \\dfrac{4+3+5}{2} - 3 = 3$. Значит $AK = 3$ см. +$$MK = AK - AM = 3 - 2 = 1\\text{ см}$$ +
Ответ: $MK = 1$ см
` + }, + { + text: `Упростите выражение $\\dfrac{|x-2| + |x+4| - |x|}{|x+2|}$ при $x < -4$.`, + sol: `Определение модуля: $|a| = a$ при $a \\geq 0$ и $|a| = -a$ при $a \\lt 0$. +
Метод раскрытия модулей: определяем знак подмодульного выражения при заданном условии на $x$, после чего знак модуля убираем (если выражение отрицательно — с противоположным знаком). +
Шаг 1. Определим знаки подмодульных выражений при $x \\lt -4$. Так как $x \\lt -4$, то: +
    +
  • $x - 2 \\lt -4 - 2 = -6 \\lt 0$, значит $|x-2| = -(x-2) = 2 - x$;
    • +
  • $x + 4 \\lt 0$ (так как $x \\lt -4$), значит $|x+4| = -(x+4) = -x - 4$;
    • +
  • $x \\lt -4 \\lt 0$, значит $|x| = -x$;
    • +
  • $x + 2 \\lt -4 + 2 = -2 \\lt 0$, значит $|x+2| = -(x+2) = -x - 2$.
    • +
+Шаг 2. Упростим числитель, подставив раскрытые модули: +$$|x-2| + |x+4| - |x| = (2-x) + (-x-4) - (-x).$$ +Раскроем скобки (минус перед $(-x)$ меняет знак): +$$= 2 - x - x - 4 + x = -2 - x = -(x+2).$$ +Шаг 3. Запишем знаменатель: $|x+2| = -(x+2)$. +
Шаг 4. Сократим дробь. При $x \\lt -4$ значение $x + 2 \\neq 0$, поэтому сокращение допустимо: +$$\\dfrac{-(x+2)}{-(x+2)} = 1.$$ +
Ответ: $1$
` + }, + ] +}; diff --git a/frontend/js/exam9/variants/v80.js b/frontend/js/exam9/variants/v80.js new file mode 100644 index 0000000..85a8477 --- /dev/null +++ b/frontend/js/exam9/variants/v80.js @@ -0,0 +1,130 @@ +VARIANTS[80] = { + label: "Вариант 80", + tasks: [ + { + text: `Определите наименьшее из значений числовых выражений:`, + opts: [ + ["а", "$3^2$"], ["б", "$3^{-2}$"], ["в", "$3^{-1}$"], + ["г", "$(-3)^0$"], ["д", "$\\left(\\dfrac{1}{8}\\right)^{-1}$"], + ], + sol: `а)$9$; б)$\\dfrac{1}{9}$; в)$\\dfrac{1}{3}$; г)$1$; д)$8$. Наименьшее $\\dfrac{1}{9}$. +
Ответ: б) $3^{-2}=\\dfrac{1}{9}$
` + }, + { + text: `Наименьшим целым решением неравенства $2x > -3$ является:`, + opts: [ + ["а", "$1$"], ["б", "$-1{,}5$"], ["в", "$-2$"], ["г", "$-1$"], ["д", "$0$"], + ], + sol: `$2x > -3 \\implies x > -1{,}5$. Наименьшее целое, строго большее $-1{,}5$: $x=-1$. +
Ответ: г) $-1$
` + }, + { + text: `Какое из следующих утверждений НЕ верно:`, + opts: [ + ["а", "площадь параллелограмма можно найти по формуле $S = a h_a$;"], + ["б", "если диагонали трапеции равны, то она — равнобедренная;"], + ["в", "$\\operatorname{tg} 45^{\\circ} = 1$;"], + ["г", "окружность, вписанная в четырёхугольник, проходит через все его вершины?"], + ], + sol: `а) верно; б) верно; в) верно. +
г) НЕВЕРНО — вписанная окружность касается сторон, а не проходит через вершины (через вершины проходит описанная). +
Ответ: г)
` + }, + { + text: `Упростите выражение $\\dfrac{t^2-b^2}{t-b} - (2t+b)$ + и найдите его значение при $t = -12$.`, + sol: `$\\dfrac{(t-b)(t+b)}{t-b}-(2t+b)=(t+b)-(2t+b)=-t$ (при $t\\neq b$). +
При $t=-12$: $-(-12)=12$. +
Ответ: $-t$;  при $t=-12$ значение равно $12$
` + }, + { + text: `При каком значении $x$ числа $x-4$, $2x-4$, $5x+2$ + являются последовательными членами арифметической прогрессии?`, + sol: `Характеристическое свойство арифметической прогрессии: каждый член (начиная со второго) равен среднему арифметическому соседних. Если $a$, $b$, $c$ — три последовательных члена АП, то $2b = a + c$. +
Шаг 1. Применим свойство для $a = x-4$, $b = 2x-4$, $c = 5x+2$: +$$2(2x - 4) = (x - 4) + (5x + 2).$$ +Шаг 2. Раскроем скобки и приведём подобные: +$$4x - 8 = 6x - 2.$$ +Шаг 3. Перенесём $x$-ы в одну часть, числа в другую: +$$4x - 6x = -2 + 8,$$ +$$-2x = 6.$$ +Шаг 4. Разделим на $-2$: +$$x = -3.$$ +Проверка. При $x = -3$ члены: $x-4 = -7$, $2x-4 = -10$, $5x+2 = -13$. Разности: $-10-(-7) = -3$, $-13-(-10) = -3$ — равны, прогрессия ✓. +
Ответ: $x=-3$
` + }, + { + text: `Найдите площадь треугольника $ABC$, если размеры одной клетки $1$ см $\\times$ $1$ см.`, + figure: ``, + sol: `Формула площади треугольника по координатам вершин (формула «шнурков»): +$$S = \\dfrac{1}{2}\\bigl|x_{A}(y_{B}-y_{C}) + x_{B}(y_{C}-y_{A}) + x_{C}(y_{A}-y_{B})\\bigr|.$$ +Альтернатива — метод «описанного прямоугольника»: описать вокруг треугольника прямоугольник со сторонами по линиям клеток; его площадь подсчитать по клеткам, а затем вычесть площади трёх прямоугольных треугольников, отсекаемых по углам. +
Шаг 1. По рисунку определить координаты вершин $A$, $B$, $C$ в клетках. +
Шаг 2. Подставить координаты в формулу или применить метод описанного прямоугольника. +
Шаг 3. Поскольку клетка имеет размер $1\\times 1$ см, площадь сразу получается в см². +
Ответ: определяется по рисунку
` + }, + { + text: `Из всех учащихся, участвующих в спортивных соревнованиях, + семиклассников было $8$, учащихся восьмых классов — $10$, девятых — $12$, + десятых — $14$, одиннадцатиклассников — $16$. + Какой процент всех участников составили учащиеся X–XI классов?`, + sol: `Формула вычисления процентного отношения: чтобы найти, какой процент составляет число $a$ от числа $b$, надо отношение $\\dfrac{a}{b}$ умножить на $100\\%$. +
Шаг 1. Найдём общее число участников, сложив количество учащихся всех классов: +$$N = 8 + 10 + 12 + 14 + 16 = 60.$$ +Шаг 2. Найдём число учащихся X–XI классов (десятиклассников и одиннадцатиклассников): +$$N_{10-11} = 14 + 16 = 30.$$ +Шаг 3. Найдём процентное отношение: +$$\\dfrac{N_{10-11}}{N}\\cdot 100\\% = \\dfrac{30}{60}\\cdot 100\\% = \\dfrac{1}{2}\\cdot 100\\% = 50\\%.$$ +
Ответ: $50\\%$
` + }, + { + text: `Не выполняя построения, найдите координаты точек пересечения параболы + $y = -x^2 + 3$ и прямой $y = -2x - 5$.`, + sol: `Метод поиска точек пересечения графиков: в точках пересечения значения функций совпадают — приравниваем правые части. +
Теорема Виета (обратная): если $x_{1}+x_{2}=-p$ и $x_{1}\\cdot x_{2}=q$, то $x_{1}$, $x_{2}$ — корни уравнения $x^{2}+px+q=0$. +
Шаг 1. Приравняем правые части: +$$-x^2 + 3 = -2x - 5.$$ +Шаг 2. Перенесём всё в левую часть и приведём к стандартному виду. Удобнее сразу умножить на $-1$, чтобы коэффициент при $x^2$ стал положительным: +$$x^2 - 2x - 8 = 0.$$ +Шаг 3. По теореме Виета ищем два числа, у которых сумма $2$, произведение $-8$. Подходят $4$ и $-2$. Значит, $(x-4)(x+2) = 0$, откуда $x_{1} = 4$, $x_{2} = -2$. +
Шаг 4. Найдём ординаты, подставив корни в уравнение прямой $y = -2x - 5$: +$$\\text{при } x = 4:\\;\\; y = -2\\cdot 4 - 5 = -8 - 5 = -13;$$ +$$\\text{при } x = -2:\\;\\; y = -2\\cdot(-2) - 5 = 4 - 5 = -1.$$ +
Ответ: $(4;\\,-13)$ и $(-2;\\,-1)$
` + }, + { + text: `В прямоугольнике $ABCD$ $AB = 5$ см, $AD = 12$ см. В треугольники $ABC$ и $ADC$ + вписаны окружности, которые касаются диагонали $AC$ в точках $M$ и $K$. + Найдите длину отрезка $MK$.`, + figure: ``, + sol: `Диагональ $AC = \\sqrt{AB^2+BC^2} = \\sqrt{25+144} = \\sqrt{169} = 13$ см. +
Оба треугольника $ABC$ и $ACD$ прямоугольные с катетами $5$ и $12$, гипотенузой $13$. +
Шаг 1. Радиус вписанной окружности: $r = \\dfrac{5+12-13}{2} = 2$ см +
Шаг 2. В $\\triangle ABC$: касательная из $A$ = $s - BC = \\dfrac{5+12+13}{2} - 12 = 3$ см, $AM = 3$ см. +
Шаг 3. В $\\triangle ACD$: касательная из $A$ = $s - CD = \\dfrac{12+5+13}{2} - 5 = 10$ см, $AK = 10$ см. +$$MK = AK - AM = 10 - 3 = 7\\text{ см}$$ +
Ответ: $MK = 7$ см
` + }, + { + text: `Упростите выражение $\\dfrac{|x-1| + |x+3| - |x-4|}{|2x+12|}$ при $x < -6$.`, + sol: `Определение модуля: $|a| = a$ при $a \\geq 0$ и $|a| = -a$ при $a \\lt 0$. +
Метод раскрытия модулей: для каждого выражения под модулем определяем знак при заданном условии на $x$, после чего снимаем модуль. +
Шаг 1. Определим знаки подмодульных выражений при $x \\lt -6$. +
    +
  • $x - 1 \\lt -6 - 1 = -7 \\lt 0$, значит $|x-1| = -(x-1) = 1 - x$;
    • +
  • $x + 3 \\lt -6 + 3 = -3 \\lt 0$, значит $|x+3| = -(x+3) = -x - 3$;
    • +
  • $x - 4 \\lt -6 - 4 = -10 \\lt 0$, значит $|x-4| = -(x-4) = 4 - x$;
    • +
  • $2x + 12 = 2(x + 6) \\lt 2\\cdot 0 = 0$ (так как $x + 6 \\lt 0$ при $x \\lt -6$), значит $|2x+12| = -(2x+12) = -2(x+6)$.
    • +
+Шаг 2. Упростим числитель, подставив раскрытые модули: +$$|x-1| + |x+3| - |x-4| = (1-x) + (-x-3) - (4-x).$$ +Раскроем скобки (минус перед скобкой меняет знак): +$$= 1 - x - x - 3 - 4 + x = -6 - x = -(x+6).$$ +Шаг 3. Запишем знаменатель: $|2x+12| = -2(x+6)$. +
Шаг 4. Сократим дробь. При $x \\lt -6$ имеем $x + 6 \\neq 0$, поэтому сокращение возможно: +$$\\dfrac{-(x+6)}{-2(x+6)} = \\dfrac{1}{2}.$$ +
Ответ: $\\dfrac{1}{2}$
` + }, + ] +}; diff --git a/js/api.js b/js/api.js index 0db9811..382c132 100644 --- a/js/api.js +++ b/js/api.js @@ -597,6 +597,7 @@ async function hideDisabledFeatures() { board: ['/board'], biochem: ['/biochem', '/biochem-library', '/biochem-reactions'], live_quiz: ['/live-quiz'], + exam9: ['/exam9', '/exam9.html'], }; for (const [key, hrefs] of Object.entries(map)) { if (feats[key] === false) { diff --git a/js/sidebar.js b/js/sidebar.js index ab0fe29..98fa395 100644 --- a/js/sidebar.js +++ b/js/sidebar.js @@ -62,6 +62,7 @@ ${L('/collection', 'layers', 'Коллекция')} ${L('/knowledge-map', 'share-2', 'Карта знаний')} ${L('/red-book', 'leaf', 'Красная книга')} + ${L('/exam9', 'clipboard-check', 'Экзамен 9 класс')} ${L('/classroom', 'presentation', 'Онлайн-урок')} ${L('/lesson-history','archive', 'Архив уроков')}

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