Решите систему уравнений: x+y-xy=-2; xy(x+y)=48.

Ответ:

\[\left\{ \begin{matrix} x + y - xy = - 2 \\ \text{xy}(x + y) = 48\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x + y = xy - 2\ \ \ \\ \text{xy}(xy - 2) = 48 \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x + y = xy - 2 \\ xy^{2} - 2xy = 48 \\ \end{matrix} \right.\ \]

\[Пусть\ \ \ xy = t:\]

\[t^{2} - 2t - 48 = 0\ \ \ \]

\[t_{1} + t_{2} = 2;\ \ \ \ \ \ \ \ \ \ \ \ t_{1} = 8\]

\[t_{1} \cdot t_{2} = - 48;\ \ \ \ \ \ \ \ t_{2} = - 6\]

\[\left\{ \begin{matrix} xy = 8\ \ \ \ \ \\ x + y = 6 \\ \end{matrix}\ \ \ \ \ или\ \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} xy = - 6\ \ \ \ \ \ \\ x + y = - 8 \\ \end{matrix} \right.\ \]

\[t^{2} - 6t + 8 = 0\ \ \ \ \ \ \]

\[t_{1} + t_{2} = 6;\ \ \ \ \ \ \ \ t_{1} = 4\ \ \ \]

\[t_{1}{\cdot t}_{2} = 8;\ \ \ \ \ \ \ \ \ \ t_{2} = 2\]

\[t^{2} + 8t - 6 = 0\]

\[D = 64 + 24 = 88\]

\[t = \frac{- 8 \pm 2\sqrt{22}}{2} = - 4 \pm \sqrt{22}.\]

\[Ответ:(4;2);\ (2;4);\ \]

\[\left( - 4 - \sqrt{22};\ - 4 + \sqrt{22} \right);\ \ \]

\[\left( - 4 + \sqrt{22};\ - 4 - \sqrt{22} \right).\]