Question

Решите систему уравнений:{x²-у²=-40{x + y = 4​ Смотреть ответ ЭЛАЙТ ЭЛАЙТ {x^2+y^2=40 {x+y=4 Во втором уравнении системы выразим Х через У: x=4-y Подставим x=4-y в первое уравнение вместо Х: {(4-y)^2+y^2=40 {x=4-y Решим первое уравнение: (4-y)^2+y^2=40 16-8y+y^2+y^2-40=0 2y^2-8y-24=0|:2 y^2-4y-12=0 D=(-4)^2-4*1*(-12)=64 y1=(4-8)/2=-2 y2=(4+8)/2=6 Находим значения Х: 1)если у=-2, то x=4-(-2)=6 2) если y=6, то x=4-6=-2 Ответ: (6;-2) (-2;6) Новые вопросы в Алгебра вычислите площадь фигуры ограниченной линиями у=х^2-3х-2 и у=2х-6 Найдите наибольшее и наименьшее значения функции на отрезке: f(x)=x^3-3x^2-9x+23; [-4;4] 60 балів !Терміново!Задача на фото) Решите неравенство 9^х-8*3^х+12>0 Решите уравнение sin^2x+2sinxcosx-3cos^2x=0 Предыдущий Следующий Вместе мы знаем больше Этот сайт использует cookies Политика Cookies . Вы можете указать условия хранения и доступ к cookies в своем браузере. 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Answer 1

Вы решили систему уравнений правильно. Вот ваш рассчет:

Система уравнений:

1. $x^2 + y^2 = 40$
2. $x + y = 4$

Из второго уравнения можно выразить $x$ через $y$: $x = 4 - y$.

Подставив $x = 4 - y$ в первое уравнение, получим: $(4 - y)^2 + y^2 = 40$ $16 - 8y + y^2 + y^2 = 40$ $2y^2 - 8y - 24 = 0$

Решим это квадратное уравнение: $y^2 - 4y - 12 = 0$ Дискриминант $D = (-4)^2 - 4 \cdot 1 \cdot (-12) = 64$

Теперь найдем корни $y_1$ и $y_2$ уравнения: $y_1 = \frac{4 - (-4)}{2} = -2$ $y_2 = \frac{4 + (-4)}{2} = 6$

Подставляем найденные значения $y$ в уравнение $x = 4 - y$ и находим соответствующие значения $x$:

1. Если $y = -2$, то $x = 4 - (-2) = 6$
2. Если $y = 6$, то $x = 4 - 6 = -2$

Итак, решение системы состоит из двух пар значений: $(6, -2)$ и $(-2, 6)$.

Что касается дополнительных вопросов:

1. Площадь фигуры ограниченной линиями $y = x^2 - 3x - 2$ и $y = 2x - 6$ можно найти, используя определенный интеграл. Формула для вычисления площади между двумя кривыми задается как: $\int_{a}^{b} |f(x) - g(x)| \, dx$, где $f(x)$ и $g(x)$ - функции, задающие кривые, а $a$ и $b$ - точки пересечения кривых. В данном случае, $f(x) = x^2 - 3x - 2$ и $g(x) = 2x - 6$, и нужно найти точки пересечения этих кривых.

2. Для нахождения наибольшего и наименьшего значения функции $f(x) = x^3 - 3x^2 - 9x + 23$ на отрезке $[-4, 4]$, следует найти критические точки (где производная равна нулю или не существует) и значения на концах интервала.

3. Решение неравенства $9^x - 8 \cdot 3^x + 12 > 0$ требует анализа степенных функций и нахождения интервалов, на которых неравенство выполняется.

4. Уравнение $\sin^2x + 2\sin x \cos x - 3\cos^2x = 0$ можно решить, используя тригонометрические тождества и замены, чтобы привести его к более простому виду.

Если у вас есть более подробные вопросы по каждому из этих пунктов, пожалуйста, задайте их, и я буду рад помочь!