Номер 87, страница 31 - гдз по алгебре 9 класс учебник Солтан, Солтан

Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019

Авторы: Солтан Г. Н., Солтан А. Е., Жумадилова А. Ж.

Тип: Учебник

Издательство: Кокшетау

Год издания: 2019 - 2026

ISBN: 978-601-317-424-2

Рекомендовано Министерством образования и науки Республики Казахстан

I. Уравнения, неравенства с двумя переменными и их системы. 2. Системы нелинейных уравнений с двумя переменными - номер 87, страница 31.

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            "text" => "<p><strong>87.</strong> Используя способ подстановки, найдите все пары (m; n) чисел, являющиеся решениями системы уравнений:</p><p><strong>a)</strong> $\begin{cases} m + n = -50, \\ mn = 96; \end{cases}$</p><p><strong>б)</strong> $\begin{cases} m^2 - n^2 = 5, \\ mn = 6; \end{cases}$</p><p><strong>в)</strong> $\begin{cases} m^2 + mn + n^2 = 13, \\ m + n = 4; \end{cases}$</p><p><strong>г)</strong> $\begin{cases} 3m - n = 0, \\ m^2 + n^2 + 2n = 9. \end{cases}$</p>"
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            "text" => "<p><strong>а)</strong></p><p>Дана система уравнений:</p><p>$\begin{cases} m + n = -50 \\ mn = 96 \end{cases}$</p><p><strong>1.</strong> Выразим <i>m</i> из первого уравнения:</p><p>$m = -50 - n$</p><p><strong>2.</strong> Подставим это выражение во второе уравнение:</p><p>$(-50 - n)n = 96$</p><p><strong>3.</strong> Раскроем скобки и решим полученное квадратное уравнение:</p><p>$-50n - n^2 = 96$</p><p>$n^2 + 50n + 96 = 0$</p><p>Для решения используем формулу корней квадратного уравнения $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.</p><p>Дискриминант $D = b^2 - 4ac = 50^2 - 4 \cdot 1 \cdot 96 = 2500 - 384 = 2116$.</p><p>Корень из дискриминанта $\sqrt{2116} = 46$.</p><p>$n_1 = \frac{-50 + 46}{2} = \frac{-4}{2} = -2$</p><p>$n_2 = \frac{-50 - 46}{2} = \frac{-96}{2} = -48$</p><p><strong>4.</strong> Найдем соответствующие значения <i>m</i>:</p><p>Если $n_1 = -2$, то $m_1 = -50 - (-2) = -48$.</p><p>Если $n_2 = -48$, то $m_2 = -50 - (-48) = -2$.</p><p>Таким образом, решениями системы являются две пары чисел.</p><p><strong>Ответ:</strong> $(-48; -2)$ и $(-2; -48)$.</p><p><strong>б)</strong></p><p>Дана система уравнений:</p><p>$\begin{cases} m^2 - n^2 = 5 \\ mn = 6 \end{cases}$</p><p><strong>1.</strong> Выразим <i>m</i> из второго уравнения (при $n \neq 0$):</p><p>$m = \frac{6}{n}$</p><p><strong>2.</strong> Подставим это выражение в первое уравнение:</p><p>$(\frac{6}{n})^2 - n^2 = 5$</p><p><strong>3.</strong> Решим полученное уравнение:</p><p>$\frac{36}{n^2} - n^2 = 5$</p><p>Умножим обе части уравнения на $n^2$ (это возможно, так как $n \neq 0$):</p><p>$36 - n^4 = 5n^2$</p><p>$n^4 + 5n^2 - 36 = 0$</p><p><strong>4.</strong> Сделаем замену переменной. Пусть $t = n^2$, где $t \ge 0$.</p><p>$t^2 + 5t - 36 = 0$</p><p>Решим это квадратное уравнение относительно <i>t</i>. По теореме Виета, корни $t_1 = 4$ и $t_2 = -9$.</p><p>Корень $t_2 = -9$ не удовлетворяет условию $t \ge 0$, поэтому его отбрасываем.</p><p><strong>5.</strong> Вернемся к переменной <i>n</i>:</p><p>$n^2 = 4$</p><p>Отсюда $n_1 = 2$ и $n_2 = -2$.</p><p><strong>6.</strong> Найдем соответствующие значения <i>m</i>:</p><p>Если $n_1 = 2$, то $m_1 = \frac{6}{2} = 3$.</p><p>Если $n_2 = -2$, то $m_2 = \frac{6}{-2} = -3$.</p><p>Получаем две пары чисел в качестве решения.</p><p><strong>Ответ:</strong> $(3; 2)$ и $(-3; -2)$.</p><p><strong>в)</strong></p><p>Дана система уравнений:</p><p>$\begin{cases} m^2 + mn + n^2 = 13 \\ m + n = 4 \end{cases}$</p><p><strong>1.</strong> Выразим <i>m</i> из второго уравнения:</p><p>$m = 4 - n$</p><p><strong>2.</strong> Подставим это выражение в первое уравнение:</p><p>$(4 - n)^2 + (4 - n)n + n^2 = 13$</p><p><strong>3.</strong> Раскроем скобки и упростим:</p><p>$(16 - 8n + n^2) + (4n - n^2) + n^2 = 13$</p><p>$16 - 8n + n^2 + 4n - n^2 + n^2 - 13 = 0$</p><p>$n^2 - 4n + 3 = 0$</p><p><strong>4.</strong> Решим полученное квадратное уравнение. По теореме Виета, корни $n_1 = 1$ и $n_2 = 3$.</p><p><strong>5.</strong> Найдем соответствующие значения <i>m</i>:</p><p>Если $n_1 = 1$, то $m_1 = 4 - 1 = 3$.</p><p>Если $n_2 = 3$, то $m_2 = 4 - 3 = 1$.</p><p>Таким образом, решениями являются две пары чисел.</p><p><strong>Ответ:</strong> $(3; 1)$ и $(1; 3)$.</p><p><strong>г)</strong></p><p>Дана система уравнений:</p><p>$\begin{cases} 3m - n = 0 \\ m^2 + n^2 + 2n = 9 \end{cases}$</p><p><strong>1.</strong> Выразим <i>n</i> из первого уравнения:</p><p>$n = 3m$</p><p><strong>2.</strong> Подставим это выражение во второе уравнение:</p><p>$m^2 + (3m)^2 + 2(3m) = 9$</p><p><strong>3.</strong> Раскроем скобки и решим полученное уравнение:</p><p>$m^2 + 9m^2 + 6m = 9$</p><p>$10m^2 + 6m - 9 = 0$</p><p><strong>4.</strong> Решим это квадратное уравнение, используя формулу для корней:</p><p>Дискриминант $D = b^2 - 4ac = 6^2 - 4 \cdot 10 \cdot (-9) = 36 + 360 = 396$.</p><p>$\sqrt{D} = \sqrt{396} = \sqrt{36 \cdot 11} = 6\sqrt{11}$.</p><p>$m = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm 6\sqrt{11}}{2 \cdot 10} = \frac{-6 \pm 6\sqrt{11}}{20} = \frac{-3 \pm 3\sqrt{11}}{10}$.</p><p>Получаем два значения для <i>m</i>:</p><p>$m_1 = \frac{-3 + 3\sqrt{11}}{10}$</p><p>$m_2 = \frac{-3 - 3\sqrt{11}}{10}$</p><p><strong>5.</strong> Найдем соответствующие значения <i>n</i> по формуле $n = 3m$:</p><p>Если $m_1 = \frac{-3 + 3\sqrt{11}}{10}$, то $n_1 = 3 \cdot (\frac{-3 + 3\sqrt{11}}{10}) = \frac{-9 + 9\sqrt{11}}{10}$.</p><p>Если $m_2 = \frac{-3 - 3\sqrt{11}}{10}$, то $n_2 = 3 \cdot (\frac{-3 - 3\sqrt{11}}{10}) = \frac{-9 - 9\sqrt{11}}{10}$.</p><p>Получаем две пары чисел в качестве решения.</p><p><strong>Ответ:</strong> $(\frac{-3 + 3\sqrt{11}}{10}; \frac{-9 + 9\sqrt{11}}{10})$ и $(\frac{-3 - 3\sqrt{11}}{10}; \frac{-9 - 9\sqrt{11}}{10})$.</p>"
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            "text" => "<p><strong>а)</strong></p><p>Дана система уравнений:</p><p>$\begin{cases} m + n = -50 \\ mn = 96 \end{cases}$</p><p><strong>1.</strong> Выразим <i>m</i> из первого уравнения:</p><p>$m = -50 - n$</p><p><strong>2.</strong> Подставим это выражение во второе уравнение:</p><p>$(-50 - n)n = 96$</p><p><strong>3.</strong> Раскроем скобки и решим полученное квадратное уравнение:</p><p>$-50n - n^2 = 96$</p><p>$n^2 + 50n + 96 = 0$</p><p>Для решения используем формулу корней квадратного уравнения $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.</p><p>Дискриминант $D = b^2 - 4ac = 50^2 - 4 \cdot 1 \cdot 96 = 2500 - 384 = 2116$.</p><p>Корень из дискриминанта $\sqrt{2116} = 46$.</p><p>$n_1 = \frac{-50 + 46}{2} = \frac{-4}{2} = -2$</p><p>$n_2 = \frac{-50 - 46}{2} = \frac{-96}{2} = -48$</p><p><strong>4.</strong> Найдем соответствующие значения <i>m</i>:</p><p>Если $n_1 = -2$, то $m_1 = -50 - (-2) = -48$.</p><p>Если $n_2 = -48$, то $m_2 = -50 - (-48) = -2$.</p><p>Таким образом, решениями системы являются две пары чисел.</p><p><strong>Ответ:</strong> $(-48; -2)$ и $(-2; -48)$.</p><p><strong>б)</strong></p><p>Дана система уравнений:</p><p>$\begin{cases} m^2 - n^2 = 5 \\ mn = 6 \end{cases}$</p><p><strong>1.</strong> Выразим <i>m</i> из второго уравнения (при $n \neq 0$):</p><p>$m = \frac{6}{n}$</p><p><strong>2.</strong> Подставим это выражение в первое уравнение:</p><p>$(\frac{6}{n})^2 - n^2 = 5$</p><p><strong>3.</strong> Решим полученное уравнение:</p><p>$\frac{36}{n^2} - n^2 = 5$</p><p>Умножим обе части уравнения на $n^2$ (это возможно, так как $n \neq 0$):</p><p>$36 - n^4 = 5n^2$</p><p>$n^4 + 5n^2 - 36 = 0$</p><p><strong>4.</strong> Сделаем замену переменной. Пусть $t = n^2$, где $t \ge 0$.</p><p>$t^2 + 5t - 36 = 0$</p><p>Решим это квадратное уравнение относительно <i>t</i>. По теореме Виета, корни $t_1 = 4$ и $t_2 = -9$.</p><p>Корень $t_2 = -9$ не удовлетворяет условию $t \ge 0$, поэтому его отбрасываем.</p><p><strong>5.</strong> Вернемся к переменной <i>n</i>:</p><p>$n^2 = 4$</p><p>Отсюда $n_1 = 2$ и $n_2 = -2$.</p><p><strong>6.</strong> Найдем соответствующие значения <i>m</i>:</p><p>Если $n_1 = 2$, то $m_1 = \frac{6}{2} = 3$.</p><p>Если $n_2 = -2$, то $m_2 = \frac{6}{-2} = -3$.</p><p>Получаем две пары чисел в качестве решения.</p><p><strong>Ответ:</strong> $(3; 2)$ и $(-3; -2)$.</p><p><strong>в)</strong></p><p>Дана система уравнений:</p><p>$\begin{cases} m^2 + mn + n^2 = 13 \\ m + n = 4 \end{cases}$</p><p><strong>1.</strong> Выразим <i>m</i> из второго уравнения:</p><p>$m = 4 - n$</p><p><strong>2.</strong> Подставим это выражение в первое уравнение:</p><p>$(4 - n)^2 + (4 - n)n + n^2 = 13$</p><p><strong>3.</strong> Раскроем скобки и упростим:</p><p>$(16 - 8n + n^2) + (4n - n^2) + n^2 = 13$</p><p>$16 - 8n + n^2 + 4n - n^2 + n^2 - 13 = 0$</p><p>$n^2 - 4n + 3 = 0$</p><p><strong>4.</strong> Решим полученное квадратное уравнение. По теореме Виета, корни $n_1 = 1$ и $n_2 = 3$.</p><p><strong>5.</strong> Найдем соответствующие значения <i>m</i>:</p><p>Если $n_1 = 1$, то $m_1 = 4 - 1 = 3$.</p><p>Если $n_2 = 3$, то $m_2 = 4 - 3 = 1$.</p><p>Таким образом, решениями являются две пары чисел.</p><p><strong>Ответ:</strong> $(3; 1)$ и $(1; 3)$.</p><p><strong>г)</strong></p><p>Дана система уравнений:</p><p>$\begin{cases} 3m - n = 0 \\ m^2 + n^2 + 2n = 9 \end{cases}$</p><p><strong>1.</strong> Выразим <i>n</i> из первого уравнения:</p><p>$n = 3m$</p><p><strong>2.</strong> Подставим это выражение во второе уравнение:</p><p>$m^2 + (3m)^2 + 2(3m) = 9$</p><p><strong>3.</strong> Раскроем скобки и решим полученное уравнение:</p><p>$m^2 + 9m^2 + 6m = 9$</p><p>$10m^2 + 6m - 9 = 0$</p><p><strong>4.</strong> Решим это квадратное уравнение, используя формулу для корней:</p><p>Дискриминант $D = b^2 - 4ac = 6^2 - 4 \cdot 10 \cdot (-9) = 36 + 360 = 396$.</p><p>$\sqrt{D} = \sqrt{396} = \sqrt{36 \cdot 11} = 6\sqrt{11}$.</p><p>$m = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm 6\sqrt{11}}{2 \cdot 10} = \frac{-6 \pm 6\sqrt{11}}{20} = \frac{-3 \pm 3\sqrt{11}}{10}$.</p><p>Получаем два значения для <i>m</i>:</p><p>$m_1 = \frac{-3 + 3\sqrt{11}}{10}$</p><p>$m_2 = \frac{-3 - 3\sqrt{11}}{10}$</p><p><strong>5.</strong> Найдем соответствующие значения <i>n</i> по формуле $n = 3m$:</p><p>Если $m_1 = \frac{-3 + 3\sqrt{11}}{10}$, то $n_1 = 3 \cdot (\frac{-3 + 3\sqrt{11}}{10}) = \frac{-9 + 9\sqrt{11}}{10}$.</p><p>Если $m_2 = \frac{-3 - 3\sqrt{11}}{10}$, то $n_2 = 3 \cdot (\frac{-3 - 3\sqrt{11}}{10}) = \frac{-9 - 9\sqrt{11}}{10}$.</p><p>Получаем две пары чисел в качестве решения.</p><p><strong>Ответ:</strong> $(\frac{-3 + 3\sqrt{11}}{10}; \frac{-9 + 9\sqrt{11}}{10})$ и $(\frac{-3 - 3\sqrt{11}}{10}; \frac{-9 - 9\sqrt{11}}{10})$.</p>"
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    "page" => Illuminate\Database\Eloquent\Collection {#1407
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        0 => App\Models\BookPage {#1032
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№87 (с. 31)
Условие. №87 (с. 31)
скриншот условия
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 31, номер 87, Условие

87. Используя способ подстановки, найдите все пары (m; n) чисел, являющиеся решениями системы уравнений:

a) $\begin{cases} m + n = -50, \\ mn = 96; \end{cases}$

б) $\begin{cases} m^2 - n^2 = 5, \\ mn = 6; \end{cases}$

в) $\begin{cases} m^2 + mn + n^2 = 13, \\ m + n = 4; \end{cases}$

г) $\begin{cases} 3m - n = 0, \\ m^2 + n^2 + 2n = 9. \end{cases}$

Решение. №87 (с. 31)
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 31, номер 87, Решение
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 31, номер 87, Решение (продолжение 2)
Решение 2 (rus). №87 (с. 31)

а)

Дана система уравнений:

$\begin{cases} m + n = -50 \\ mn = 96 \end{cases}$

1. Выразим m из первого уравнения:

$m = -50 - n$

2. Подставим это выражение во второе уравнение:

$(-50 - n)n = 96$

3. Раскроем скобки и решим полученное квадратное уравнение:

$-50n - n^2 = 96$

$n^2 + 50n + 96 = 0$

Для решения используем формулу корней квадратного уравнения $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Дискриминант $D = b^2 - 4ac = 50^2 - 4 \cdot 1 \cdot 96 = 2500 - 384 = 2116$.

Корень из дискриминанта $\sqrt{2116} = 46$.

$n_1 = \frac{-50 + 46}{2} = \frac{-4}{2} = -2$

$n_2 = \frac{-50 - 46}{2} = \frac{-96}{2} = -48$

4. Найдем соответствующие значения m:

Если $n_1 = -2$, то $m_1 = -50 - (-2) = -48$.

Если $n_2 = -48$, то $m_2 = -50 - (-48) = -2$.

Таким образом, решениями системы являются две пары чисел.

Ответ: $(-48; -2)$ и $(-2; -48)$.

б)

Дана система уравнений:

$\begin{cases} m^2 - n^2 = 5 \\ mn = 6 \end{cases}$

1. Выразим m из второго уравнения (при $n \neq 0$):

$m = \frac{6}{n}$

2. Подставим это выражение в первое уравнение:

$(\frac{6}{n})^2 - n^2 = 5$

3. Решим полученное уравнение:

$\frac{36}{n^2} - n^2 = 5$

Умножим обе части уравнения на $n^2$ (это возможно, так как $n \neq 0$):

$36 - n^4 = 5n^2$

$n^4 + 5n^2 - 36 = 0$

4. Сделаем замену переменной. Пусть $t = n^2$, где $t \ge 0$.

$t^2 + 5t - 36 = 0$

Решим это квадратное уравнение относительно t. По теореме Виета, корни $t_1 = 4$ и $t_2 = -9$.

Корень $t_2 = -9$ не удовлетворяет условию $t \ge 0$, поэтому его отбрасываем.

5. Вернемся к переменной n:

$n^2 = 4$

Отсюда $n_1 = 2$ и $n_2 = -2$.

6. Найдем соответствующие значения m:

Если $n_1 = 2$, то $m_1 = \frac{6}{2} = 3$.

Если $n_2 = -2$, то $m_2 = \frac{6}{-2} = -3$.

Получаем две пары чисел в качестве решения.

Ответ: $(3; 2)$ и $(-3; -2)$.

в)

Дана система уравнений:

$\begin{cases} m^2 + mn + n^2 = 13 \\ m + n = 4 \end{cases}$

1. Выразим m из второго уравнения:

$m = 4 - n$

2. Подставим это выражение в первое уравнение:

$(4 - n)^2 + (4 - n)n + n^2 = 13$

3. Раскроем скобки и упростим:

$(16 - 8n + n^2) + (4n - n^2) + n^2 = 13$

$16 - 8n + n^2 + 4n - n^2 + n^2 - 13 = 0$

$n^2 - 4n + 3 = 0$

4. Решим полученное квадратное уравнение. По теореме Виета, корни $n_1 = 1$ и $n_2 = 3$.

5. Найдем соответствующие значения m:

Если $n_1 = 1$, то $m_1 = 4 - 1 = 3$.

Если $n_2 = 3$, то $m_2 = 4 - 3 = 1$.

Таким образом, решениями являются две пары чисел.

Ответ: $(3; 1)$ и $(1; 3)$.

г)

Дана система уравнений:

$\begin{cases} 3m - n = 0 \\ m^2 + n^2 + 2n = 9 \end{cases}$

1. Выразим n из первого уравнения:

$n = 3m$

2. Подставим это выражение во второе уравнение:

$m^2 + (3m)^2 + 2(3m) = 9$

3. Раскроем скобки и решим полученное уравнение:

$m^2 + 9m^2 + 6m = 9$

$10m^2 + 6m - 9 = 0$

4. Решим это квадратное уравнение, используя формулу для корней:

Дискриминант $D = b^2 - 4ac = 6^2 - 4 \cdot 10 \cdot (-9) = 36 + 360 = 396$.

$\sqrt{D} = \sqrt{396} = \sqrt{36 \cdot 11} = 6\sqrt{11}$.

$m = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm 6\sqrt{11}}{2 \cdot 10} = \frac{-6 \pm 6\sqrt{11}}{20} = \frac{-3 \pm 3\sqrt{11}}{10}$.

Получаем два значения для m:

$m_1 = \frac{-3 + 3\sqrt{11}}{10}$

$m_2 = \frac{-3 - 3\sqrt{11}}{10}$

5. Найдем соответствующие значения n по формуле $n = 3m$:

Если $m_1 = \frac{-3 + 3\sqrt{11}}{10}$, то $n_1 = 3 \cdot (\frac{-3 + 3\sqrt{11}}{10}) = \frac{-9 + 9\sqrt{11}}{10}$.

Если $m_2 = \frac{-3 - 3\sqrt{11}}{10}$, то $n_2 = 3 \cdot (\frac{-3 - 3\sqrt{11}}{10}) = \frac{-9 - 9\sqrt{11}}{10}$.

Получаем две пары чисел в качестве решения.

Ответ: $(\frac{-3 + 3\sqrt{11}}{10}; \frac{-9 + 9\sqrt{11}}{10})$ и $(\frac{-3 - 3\sqrt{11}}{10}; \frac{-9 - 9\sqrt{11}}{10})$.

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