ГДЗ по алгебре 9 класс Макарычев Задание 508

\[\boxed{\text{508\ (508).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[\Longrightarrow \left\{ \begin{matrix} (x + 2y - 1)(x - y) = 0 \\ x² + y² = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ \left\{ \begin{matrix} x + 2y - 1 = 0 \\ x^{2} + y^{2} = 8\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 1 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ (1 - 2y)^{2} + y^{2} = 8 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 1 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \\ 5y² - 4y - 7 = 0 \\ \end{matrix} \right.\ \]

\[5y^{2} - 4y - 7 = 0\]

\[D = 4 + 35 = 39\]

\[y_{1,2} = \frac{2 \pm \sqrt{39}}{5};\]

\[\left\{ \begin{matrix} y_{1} = \frac{2 + \sqrt{39}}{5} \\ x_{1} = \frac{1 - 2\sqrt{39}}{5}\ \\ \end{matrix} \right.\ \text{\ \ }\]

\[или\ \ \left\{ \begin{matrix} y_{2} = \frac{2 - \sqrt{39}}{5} \\ x_{2} = \frac{1 + 2\sqrt{39}}{5}. \\ \end{matrix} \right.\ \]

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

\[\Longrightarrow \left\{ \begin{matrix} x = y\ \ \ \ \ \ \ \ \ \ \\ x^{2} + x^{2} = 8 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = y \\ x^{2} = 4 \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} - 6xy + 5y^{2} - x + 5y = 0 \\ x^{2} - 20y^{2} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x - y - 1)(x - 5y) = 0 \\ x² - 20y^{2} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ \left\{ \begin{matrix} x - y - 1 = 0 \\ x^{2} - {20y}^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 2y + 1 - 20y^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 19y² - 2y + 4 = 0 \\ \end{matrix} \right.\ \]

\[19y^{2} - 2y + 4 = 0\]

\[D = 1 - 19 \cdot 4 < 0 \Longrightarrow корней\ \]

\[нет;\]

\[2)\ \left\{ \begin{matrix} x - 5y = 0\ \ \ \ \ \\ x^{2} - 20y^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 5y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 25y^{2} - 20y^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 5y \\ y^{2} = 1\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = \pm 1 \\ x = 5y \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 1 \\ x_{1} = 5 \\ \end{matrix} \right.\ \text{\ \ \ }или\left\{ \begin{matrix} y_{2} = - 1 \\ x_{2} = - 5. \\ \end{matrix} \right.\ \]

\[Ответ:а)\ ( - 2;\ - 2);(2;2);\]

\[\left( \frac{1 + 2\sqrt{39}}{5};\ \frac{2 - \sqrt{39}}{5} \right);\]

\[\left( \frac{1 - 2\sqrt{39}}{5};\ \frac{2 + \sqrt{39}}{5} \right);\]

\[\ б)\ ( - 5;\ - 1);(5;1).\]