Решебник по алгебре 9 класс Мерзляк Задание 452

\[\boxed{\text{452\ (452).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \left\{ \begin{matrix} x - y = 3 \\ xy = 28\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 3 + y\ \ \ \ \ \ \ \ \\ (3 + y)y = 28 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 3 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3y + y^{2} - 28 = 0 \\ \end{matrix} \right.\ \]

\[y^{2} + 3y - 28 = 0\]

\[y_{1} + y_{2} = - 3,\ \ y_{1} = - 7\]

\[y_{1}y_{2} = - 28,\ \ y_{2} = 4\]

\[\left\{ \begin{matrix} x = - 4 \\ y = - 7 \\ \end{matrix} \right.\ \ \ \ или\ \ \left\{ \begin{matrix} x = 7 \\ y = 4 \\ \end{matrix} \right.\ \]

\[Ответ:( - 4;\ - 7);\ (7;4).\]

\[2)\ \left\{ \begin{matrix} y^{2} - x = 14 \\ x - y = - 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = - 14 + y^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ - 14 + y^{2} - y + 2 = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = y^{2} - 14\ \ \ \ \ \ \ \ \\ y^{2} - y - 12 = 0 \\ \end{matrix} \right.\ \]

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

\[y_{1} + y_{2} = 1,\ \ y_{1} = 4\]

\[y_{1}y_{2} = - 12,\ \ y_{2} = - 3\]

\[\left\{ \begin{matrix} x = 2 \\ y = 4 \\ \end{matrix} \right.\ \ \ \ или\ \text{\ \ }\left\{ \begin{matrix} x = - 5 \\ y = - 3 \\ \end{matrix} \right.\ \]

\[Ответ:(2;4);\ ( - 5;\ - 3).\]

\[3)\ \left\{ \begin{matrix} y - 2x^{2} = 2 \\ 3x + y = 1\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = 2 + 2x^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 3x + 2 + 2x^{2} - 1 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = 2 + 2x^{2}\text{\ \ \ \ \ \ \ \ \ } \\ 2x^{2} + 3x + 1 = 0 \\ \end{matrix} \right.\ \]

\[2x^{2} + 3x + 1 = 0\]

\[D = 1\]

\[x_{1} = \frac{- 3 + 1}{4} = - 0,5\]

\[x_{2} = \frac{- 3 - 1}{4} = - 1\]

\[\left\{ \begin{matrix} x = 2,5 \\ y = - 0,5 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \left\{ \begin{matrix} y = 4 \\ x = - 1 \\ \end{matrix} \right.\ \]

\[Ответ:( - 0,5;2,5);\ ( - 1;4).\]

\[4)\ \left\{ \begin{matrix} x^{2} - 2y^{2} = 8 \\ x + y = 6\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} - 2y^{2} = 8 \\ x = 6 - y\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} (6 - y)^{2} - 2y^{2} - 8 = 0 \\ x = 6 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 36 - 12y + y^{2} - 2y^{2} - 8 = 0 \\ x = 6 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[- y^{2} - 12y + 28 = 0\]

\[y_{1} + y_{2} = - 12,\ \ y_{1} = 2\]

\[y_{1}y_{2} = - 28,\ \ y_{2} = - 14\]

\[\left\{ \begin{matrix} x = 4 \\ y = 2 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \left\{ \begin{matrix} x = 20\ \ \\ y = - 14 \\ \end{matrix} \right.\ \]

\[Ответ:(4;2);\ (20;\ - 14).\]