ГДЗ по алгебре 9 класс Макарычев Задание 430

\[\boxed{\text{430\ (430).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left\{ \begin{matrix} x = 3 - y\ \ \ \ \\ y^{2} - x = 39 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 3 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 3 + y - 39 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 3 - y\ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + y - 42 = 0 \\ \end{matrix} \right.\ \]

\[y^{2} + y - 42 = 0\]

\[y_{1} + y_{2} = - 1;\ \ \ y_{1} \cdot y_{2} = - 42\]

\[y_{1} = - 7;\ \ y_{2} = 6.\]

\[1)\ \left\{ \begin{matrix} y_{1} = - 7 \\ x_{1} = 10 \\ \end{matrix} \right.\ \text{\ \ }\]

\[2)\left\{ \begin{matrix} y_{2} = 6\ \ \ \\ x_{2} = - 3 \\ \end{matrix} \right.\ .\]

\[\textbf{б)}\ \left\{ \begin{matrix} y = 1 + x\ \ \ \ \ \\ x + y^{2} = - 1 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 1 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x + (1 + x)^{2} = - 1 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 1 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x + 1 + 2x + x^{2} + 1 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 1 + x\ \ \ \ \ \ \ \ \ \\ x^{2} + 3x + 2 = 0 \\ \end{matrix} \right.\ \]

\[x^{2} + 3x + 2 = 0\]

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

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

\[1)\ \left\{ \begin{matrix} x_{1} = - 1 \\ y_{1} = 0\ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]

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

\[\textbf{в)}\ \left\{ \begin{matrix} x^{2} + y = 14 \\ y - x = 8\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = x + 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + x + 8 - 14 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = x + 8\ \ \ \ \ \ \ \ \ \\ x^{2} + x - 6 = 0 \\ \end{matrix} \right.\ \]

\[x^{2} + x - 6 = 0\]

\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 6\]

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

\[1)\ \left\{ \begin{matrix} x_{1} = 2\ \ \\ y_{1} = 10 \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} x_{2} = - 3 \\ y_{2} = 5\ \ \ \\ \end{matrix} \right.\ .\]

\[\textbf{г)}\ \left\{ \begin{matrix} x + y = 4\ \ \\ y + xy = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ y + y(4 - y) = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y + 4y - y^{2} - 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 5y + 6 = 0 \\ \end{matrix} \right.\ \]

\[y^{2} - 5y + 6 = 0\]

\[y_{1} + y_{2} = 5;\ \ \ y_{1} \cdot y_{2} = 6\]

\[y_{1} = 2;\ \ \ y_{2} = 3.\]

\[1)\ \left\{ \begin{matrix} y_{1} = 2 \\ x_{1} = 2 \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} y_{2} = 3 \\ x_{2} = 1. \\ \end{matrix} \right.\ \]

\[Ответ:а)\ (10;\ - 7);( - 3;6);\]

\[\textbf{б)}\ ( - 1;0);( - 2;\ - 1);\]

\[\textbf{в)}\ (2;10);\ \ ( - 3;5);\]

\[\textbf{г)}\ (2;2);(1;3).\]