ГДЗ по алгебре 9 класс Макарычев Задание 507

\[\boxed{\text{507\ (507)\ .}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} (x - 2y)(x + 3y) = 0 \\ x² - y^{2} = 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ \left\{ \begin{matrix} x - 2y = 0\ \ \ \\ x^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]

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

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

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 2 \\ x_{1} = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ }или\ \ \ \left\{ \begin{matrix} y_{2} = - 2 \\ x_{2} = - 4 \\ \end{matrix} \right.\ ;\]

\[2)\ \left\{ \begin{matrix} x + 3y = 0\ \ \\ x^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = - 3y\ \ \ \ \ \ \ \ \ \\ 9y^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = - 3y \\ y^{2} = \frac{3}{2}\text{\ \ \ } \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = \pm \frac{\sqrt{6}}{2} \\ x = - 3y \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = \frac{\sqrt{6}}{2}\text{\ \ \ \ \ \ \ } \\ x_{1} = - \frac{3\sqrt{6}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }или\ \]

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

\[\Longrightarrow \left\{ \begin{matrix} (x - y + 2)(x - 3y) = 0 \\ x² - xy + y^{2} = 7\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ \left\{ \begin{matrix} x - y + 2 = 0\ \ \ \ \ \\ x^{2} - xy + y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 1 \\ y_{1} = 3 \\ \end{matrix} \right.\ \text{\ \ }или\ \ \left\{ \begin{matrix} x_{2} = - 3 \\ y_{2} = - 1. \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} x - 3y = 0\ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - xy + y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 9y^{2} - 3y^{2} + y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 3y \\ y^{2} = 1\ \\ \end{matrix} \Longrightarrow \right.\ \left\{ \begin{matrix} y_{1} = 1 \\ x_{1} = 3 \\ \end{matrix} \right.\ \text{\ \ }\]

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

\[Ответ:а)\ (4;2);( - 4;\ - 2);\]

\[\left( - \frac{3\sqrt{6}}{2};\frac{\sqrt{6}}{2} \right);\left( \frac{3\sqrt{6}}{2};\ - \frac{\sqrt{6}}{2} \right);\ \]

\[\textbf{б)}\ (1;3);( - 3;\ - 1);(3;1).\]