Решебник по алгебре 9 класс Макарычев Задание 494

\[\boxed{\text{494\ (494).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\frac{x - 1}{x + 2} - \frac{1 - x}{x^{2} + 3x + 2} =\]

\[= \frac{x - 1}{x + 2} - \frac{1 - x}{(x + 2)(x + 1)} =\]

\[= \frac{(x - 1)(x + 1) - 1 + x}{(x + 1)(x + 2)} =\]

\[= \frac{x^{2} - 1 - 1 + x}{(x + 1)(x + 2)} =\]

\[= \frac{x^{2} + x - 2}{(x + 1)(x + 2)};\]

\[x^{2} + 3x + 2 = (x + 2)(x + 1)\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = 2\]

\[x_{1} = - 2;\ \ \ x_{2} = - 1.\]

\[x^{2} + x - 2 = (x + 2)(x - 1)\]

\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 2\]

\[x_{1} = - 2;\ \ x_{2} = 1.\]

\[\Longrightarrow \frac{(x + 2)(x - 1)}{(x + 1)(x + 2)} = \frac{x - 1}{x + 1}.\]

\[\boxed{\text{494.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 40 \\ xy = - 12\ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} + 2xy + y^{2} = 40 - 24 \\ 2xy = - 24\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x + y)^{2} = 16 \\ xy = - 12\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x + y = 4 \\ xy = - 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y(4 - y) + 12 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 4y - 12 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 6\ \ \ \\ x_{1} = - 2 \\ \end{matrix} \right.\ \ \ или\ \left\{ \begin{matrix} y_{2} = - 2 \\ x_{2} = 6.\ \ \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} x + y = - 4 \\ xy = - 12\ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = - 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ ( - 4 - y) \cdot y = - 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = - 4 - y\ \ \ \ \ \ \ \ \ \\ y^{2} + 4y - 12 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = - 6 \\ x_{1} = 2\ \ \ \\ \end{matrix} \right.\ \ \ \ \ или\ \left\{ \begin{matrix} y_{2} = 2\ \ \ \\ x_{2} = - 6. \\ \end{matrix} \right.\ \ \]

\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + 2y^{2} = 228\ \ \ \\ 3x^{2} - 2y^{2} = 172 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 4x^{2} = 228 + 172 \\ 2y^{2} = 228 - x^{2}\text{\ \ } \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x² = 100 \\ y² = 64\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 10 \\ y = \pm 8.\ \ \\ \end{matrix} \right.\ \]

\[Ответ:а)\ ( - 2;6);(6;\ - 2);\ \ \ \]

\[\textbf{б)}\ (10;8);( - 10;\ - 8);\]

\[(10;\ - 8);( - 10;8).\]