ГДЗ по алгебре 9 класс Макарычев Задание 511

\[\boxed{\text{511\ (511).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} \frac{x}{y} + \frac{y}{x} = \frac{25}{12} \\ x^{2} - y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 12 \cdot \left( \frac{x}{y} \right)^{2} - 25 \cdot \frac{x}{y} + 12 = 0 \\ x² - y^{2} = 7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Пусть\frac{x}{y} = t,\ \]

\[тогда\ \ 12t^{2} - 25t + 12 = 0,\]

\[D = 25^{2} - 4 \cdot 12 \cdot 12 =\]

\[= 625 - 576 = 49,\]

\[t_{1,2} = \frac{25 \pm 7}{24} = \frac{4}{3};\frac{3}{4};\ \]

\[1)\frac{x}{y} = \frac{4}{3} \Longrightarrow 3x = 4y;\]

\[\left\{ \begin{matrix} 3x = 4y\ \ \ \ \ \\ x^{2} - y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = \frac{4}{3}\text{y\ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{16}{9}y^{2} - y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = \frac{4}{3}\text{y\ \ } \\ \frac{7}{9}y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \frac{4}{3}y \\ y^{2} = 9\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[\textbf{б)}\ \left\{ \begin{matrix} \frac{x}{y} - \frac{y}{x} = 2,1\ \\ x^{2} + y^{2} = 29 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 10 \cdot \left( \frac{x}{y} \right)^{2} - 21 \cdot \frac{x}{y} - 10 = 0 \\ x² + y² = 29\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Пусть\ \frac{x}{y} = t,\ \]

\[тогда\ 10t^{2} - 21t - 10 = 0,\]

\[D = 21^{2} + 4 \cdot 10 \cdot 10 =\]

\[= 441 + 400 = 841 = 29^{2},\]

\[t_{1,2} = \frac{21 \pm 29}{20} = 2,5;\ - 0,4;\]

\[1)\frac{x}{y} = 2,5 \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 2,5y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 6,25y^{2} + y^{2} = 29 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 2,5y \\ y^{2} = 4\ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[2)\frac{x}{y} = - 0,4 \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = - 0,4y\ \ \ \ \ \ \ \ \ \ \ \ \\ 0,16y^{2} + y^{2} = 29 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y^{2} = 25\ \ \ \ \\ x = - 0,4y \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[Ответ:а)\ ( - 4;\ - 3);(4;3);\]

\[\textbf{б)}\ ( - 2;5);(2;\ - 5);(5;2);\]

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