Решебник по алгебре 9 класс Макарычев Задание 974

 

\[\boxed{\text{974\ (974).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x + xy + y = 11 \\ x - xy + y = 1\ \ \\ \end{matrix} \right.\ ( - )\]

\[x^{2} - 6x + 5 = 0\]

\[D = 36 - 20 = 16\]

\[x_{1} = \frac{6 + 4}{2} = 5,\ \ \]

\[x_{2} = \frac{6 - 4}{2} = 1,\]

\[1)\ x_{1} = 5,\ \ y_{1} = \frac{5}{5} = 1,\]

\[2)\ x_{2} = 1,\ \ y_{2} = \frac{5}{1} = 5.\]

\[Ответ:(5;1)\ или\ \ (1;5).\]

\[\textbf{б)}\ \left\{ \begin{matrix} 2x - y - xy = 14\ \\ x + 2y + xy = - 7 \\ \end{matrix} \right.\ ( + )\]

\[\Longrightarrow \left\{ \begin{matrix} y = 7 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 3x^{2} + 2x + 21 = 0 \\ \end{matrix} \right.\ \]

\[3x² - 2x - 21 = 0\]

\[D = 4 + 252 = 256\]

\[x_{1} = \frac{2 + 16}{6} = 3,\ \ \]

\[x_{2} = \frac{2 - 16}{6} = - \frac{14}{6} = - 2\frac{1}{3},\]

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

\[2)\ x = - 2\frac{1}{3},\ \ y = 14.\]

\[Ответ:(3;\ - 2)\ или\ \ \left( - 2\frac{1}{3};14 \right).\]

\[x^{2} - 8x + 15 = 0\]

\[D = 64 - 60 = 4\]

\[x_{1} = \frac{8 + 2}{2} = 5,\ \ \]

\[x_{2} = \frac{8 - 2}{2} = 3,\]

\[x_{1} = 3,\ \ y_{1} = 5;\]

\[x_{2} = 5,\ \ y_{2} = 3.\]

\[x^{2} + 8x + 15 = 0\]

\[D = 64 - 60 = 4\]

\[x_{1} = \frac{- 8 + 2}{2} = - 3,\ \ \]

\[x_{2} = \frac{- 8 - 2}{2} = - 5,\]

\[x_{1} = - 3,\ \ y_{1} = - 5;\]

\[x_{2} = - 5,\ \ y_{2} = - 3.\]

\[Ответ:(3;5),(\ 5;3),\ ( - 3;\ - 5),\ \]

\[( - 5;\ - 3).\]

\[x^{2} = t,\ \ t \geq 0,\]

\[t^{2} - 12t - 64 = 0\]

\[D = 144 + 256 = 400\]

\[t_{1} = \frac{12 + 20}{2} = 16,\]

\[t_{2} = \frac{12 - 20}{2} = - 4 \Longrightarrow не\ \]

\[подходит\ по\ условию.\]

\[x^{2} = 16,\ \ x = \pm 4,\]

\[1)\ x_{1} = 4,\ \ y_{1} = 2;\]

\[2)\ x_{2} = - 4,\ \ y_{2} = - 2.\]

\[Ответ:(4;2)\ или\ \ ( - 4;\ - 2).\]