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МатематикаРешебник по алгебре 9 класс Макарычев Задание 507
Задание 507
\[\boxed{\text{507\ (507)\ .}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} (x - 2y)(x + 3y) = 0 \\ x² - y^{2} = 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} x - 2y = 0\ \ \ \\ x^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y \\ y^{2} = 4 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = \pm 2 \\ x = \pm 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 2 \\ x_{1} = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ }или\ \ \ \left\{ \begin{matrix} y_{2} = - 2 \\ x_{2} = - 4 \\ \end{matrix} \right.\ ;\]
\[2)\ \left\{ \begin{matrix} x + 3y = 0\ \ \\ x^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 3y\ \ \ \ \ \ \ \ \ \\ 9y^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 3y \\ y^{2} = \frac{3}{2}\text{\ \ \ } \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = \pm \frac{\sqrt{6}}{2} \\ x = - 3y \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = \frac{\sqrt{6}}{2}\text{\ \ \ \ \ \ \ } \\ x_{1} = - \frac{3\sqrt{6}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }или\ \]
\[\ \left\{ \begin{matrix} y_{2} = - \frac{\sqrt{6}}{2} \\ x_{2} = \frac{3\sqrt{6}}{2}. \\ \end{matrix} \right.\ \]
\[\Longrightarrow \left\{ \begin{matrix} (x - y + 2)(x - 3y) = 0 \\ x² - xy + y^{2} = 7\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} x - y + 2 = 0\ \ \ \ \ \\ x^{2} - xy + y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = x + 2\ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2x - 3 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 1 \\ y_{1} = 3 \\ \end{matrix} \right.\ \text{\ \ }или\ \ \left\{ \begin{matrix} x_{2} = - 3 \\ y_{2} = - 1. \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} x - 3y = 0\ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - xy + y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 9y^{2} - 3y^{2} + y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 3y \\ y^{2} = 1\ \\ \end{matrix} \Longrightarrow \right.\ \left\{ \begin{matrix} y_{1} = 1 \\ x_{1} = 3 \\ \end{matrix} \right.\ \text{\ \ }\]
\[или\ \left\{ \begin{matrix} y_{2} = - 1 \\ x_{2} = - 3. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (4;2);( - 4;\ - 2);\]
\[\left( - \frac{3\sqrt{6}}{2};\frac{\sqrt{6}}{2} \right);\left( \frac{3\sqrt{6}}{2};\ - \frac{\sqrt{6}}{2} \right);\ \]
\[\textbf{б)}\ (1;3);( - 3;\ - 1);(3;1).\]
\[\boxed{\text{507.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[Пусть\ x - числитель\ дроби,\]
\[\ а\ y - знаменатель.\]
\[Составим\ систему\ уравнений:\]
\[\left\{ \begin{matrix} \frac{x^{2}}{y - 1} = 2 \\ \frac{x - 1}{y + 1} = \frac{1}{4} \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x^{2} = 2y - 2\ \ \ \ \\ 4x - 4 = y + 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 4x - 5\ \ \ \ \ \ \ \ \ \ \\ x^{2} = 8x - 10 - 2 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 4x - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 8x + 12 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 6\ \ \\ y_{1} = 19 \\ \end{matrix} \right.\ \ \ или\ \ \left\{ \begin{matrix} x_{2} = 2\ \\ y_{2} = 3. \\ \end{matrix} \right.\ \]
\[Ответ:\frac{6}{19}\ или\ \ \ \frac{2}{3}.\]
