ГДЗ по алгебре 9 класс Макарычев Задание 621

\[\boxed{\text{621\ (621).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} 9x² + 9y² = 13 \\ 3xy = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \frac{2}{3y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{9 \cdot 4}{9y²} + 9y² = 13 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \frac{2}{3y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 4 + 9y^{4} = 13y² \\ \end{matrix} \right.\ \]

\[9y^{4} - 13y^{2} + 4 = 0\]

\[D = 13^{2} - 4 \cdot 9 \cdot 4 = 169 - 144 = 25 \Longrightarrow y_{1,2} = \frac{13 \pm 5}{18};\]

\[1)\ y² = 1 \Longrightarrow y = \pm 1;\]

\[2)\ y² = \frac{8}{18},\ \ y² = \frac{4}{9} \Longrightarrow y = \pm \frac{2}{3}.\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 1 \\ x_{1} = \frac{2}{3} \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{2} = - 1 \\ x_{2} = - \frac{2}{3} \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{3} = \frac{2}{3} \\ x_{3} = 1\ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{4} = - \frac{2}{3} \\ x_{4} = - 1. \\ \end{matrix} \right.\ \]

\[Ответ:\left( \frac{2}{3};1 \right);\ \ \left( - \frac{2}{3}; - 1 \right);\ \ \left( 1;\frac{2}{3} \right);\ \ \left( - 1;\ - \frac{2}{3} \right).\]

\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 29 \\ y^{2} - 4x^{2} = 9 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y^{2} = 4x^{2} + 9\ \ \ \ \ \ \ \ \\ x^{2} + 4x^{2} + 9 = 29 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} 5x^{2} = 20\ \ \ \ \ \ \\ y^{2} = 4x^{2} + 9 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x² = 4\ \ \ \ \ \ \ \ \ \ \ \\ y² = 16 + 9 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 2 \\ y = \pm 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 2 \\ x_{1} = 5 \\ \end{matrix} \right.\ ,\ \ \left\{ \begin{matrix} y_{2} = 2\ \ \\ x_{2} = - 5 \\ \end{matrix} \right.\ ,\ \ \left\{ \begin{matrix} y_{3} = - 2 \\ x_{3} = 5\ \ \ \\ \end{matrix} \right.\ ,\ \ \left\{ \begin{matrix} y_{4} = - 2 \\ x_{4} = - 5. \\ \end{matrix} \right.\ \]

\[Ответ:(5;2);\ \ ( - 5;2);\ \ (5;\ - 2);\ \ ( - 5;\ - 2).\]