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МатематикаРешебник по алгебре 9 класс Макарычев Задание 495
Задание 495
\[\boxed{\text{495\ (495).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\left\{ \begin{matrix} 5x - y - 2 = 0\ \ \ \ \ \ \ \\ x^{2} - 2xy + y^{2} = 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 5x - 2\ \ \\ (x - y)^{2} = 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 5x - 2\ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 5x + 2)^{2} = 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 5x - 2\ \ \ \ \ \ \ \ \ \\ ( - 4x + 2)^{2} = 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 5x - 2\ \ \ \ \ \\ (4x - 2)^{2} = 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 5x - 2\ \ \ \ \\ (2x - 1)^{2} = 1 \\ \end{matrix} \right.\ \]
\[2x - 1 = 1\ \ \ \ \ \ \ \ \ 2x - 1 = - 1\]
\[2x = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x = 0\]
\[x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 1 \\ y_{1} = 3 \\ \end{matrix} \right.\ \text{\ \ }или\ \ \ \left\{ \begin{matrix} x_{2} = 0\ \ \ \ \\ y_{2} = - 2. \\ \end{matrix} \right.\ \]
\[Ответ:(1;3);\ \ (0; - 2).\]
\[\boxed{\text{495.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + 3x - 4y = 20 \\ x^{2} - 2x + y = - 5\ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x^{2} - 3x - 4y = 20 \\ - x^{2} + 2x - y = 5\ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 5x - 5y = 25 \\ 5x^{2} - 5x = 0\ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 5 + y\ \ \ \ \ \\ x(x - 1) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 0 \\ y_{1} = 5 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \left\{ \begin{matrix} x_{2} = 1\ \ \ \ \\ y_{2} = - 4. \\ \end{matrix} \right.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} y^{2} + 3x - y = 1 \\ y^{2} + 6x - 2y = 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 3x - y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2y^{2} + 6x - 2y = 2\ \ \ \\ - y^{2} - 6x + 2y = - 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 3x \\ y^{2} = 1\ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 1 \\ x_{1} = \frac{1}{3}\ \\ \end{matrix} \right.\ \ \ или\ \text{\ \ }\left\{ \begin{matrix} y_{2} = - 1\ \ \\ x_{2} = - \frac{1}{3}. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (0;5);(1;\ - 4);\ \ \]
\[\textbf{б)}\ \left( \frac{1}{3};1 \right);\left( - \frac{1}{3};\ - 1 \right).\]
