ГДЗ по алгебре 9 класс Макарычев Задание 528

\[\boxed{\text{528}\text{\ (528)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x - y = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 1)(y + 1) = 2xy + 3 \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[\Longrightarrow \left\{ \begin{matrix} x = y + 4\ \ \ \ \ \\ y(y + 4) = 0 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{1} = 0 \\ x_{1} = 4 \\ \end{matrix} \right.\ \text{\ \ }\]

\[или\ \ \ \left\{ \begin{matrix} y_{2} = - 4 \\ x_{2} = 0\ \ . \\ \end{matrix} \right.\ \]

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

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

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

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

\[\textbf{в)}\ \left\{ \begin{matrix} 2x - y = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x + 1)(y + 4) = 2xy - 1 \\ \end{matrix} \Longrightarrow \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} y = 2x - 5\ \ \ \ \ \ \ \ \ \\ - 2x^{2} + 11x = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 2x - 5\ \ \ \ \ \ \ \ \\ x(11 - 2x) = 0 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x_{1} = 0\ \ \ \\ y_{1} = - 5 \\ \end{matrix} \right.\ \text{\ \ }\]

\[или\ \ \left\{ \begin{matrix} x_{2} = 5,5 \\ y_{2} = 6.\ \ \ \\ \end{matrix} \right.\ \]

\[\textbf{г)}\ \left\{ \begin{matrix} x + y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 1)(y + 5) = y^{2} - 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 1 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - y(y + 5) - y^{2} + 12 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 1 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 2y^{2} - 5y + 12 = 0 \\ \end{matrix} \right.\ \]

\[2y^{2} + 5y - 12 = 0\]

\[D = 25 + 4 \cdot 2 \cdot 12 = 121\]

\[y_{1,2} = \frac{- 5 \pm 11}{4} = - 4;1,5;\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = - 4 \\ x_{1} = 5\ \ \ \ \\ \end{matrix} \right.\ \ \ или\ \ \ \ \left\{ \begin{matrix} y_{2} = 1,5\ \ \ \ \ \\ x_{2} = - 0,5. \\ \end{matrix} \right.\ \ \]

\[Ответ:а)\ (4;0);(0;\ - 4);\ \ \]

\[\textbf{б)}\ (2;3);( - 2;\ - 1);\]

\[\textbf{в)}\ (0;\ - 5);\ \ (5,5;6);\ \ \ \]

\[\textbf{г)}\ (5;\ - 4);( - 0,5;1,5).\]