Решите систему уравнений: x+y+xy=-19; xy(x+y)=-20.

\[\left\{ \begin{matrix} x + y + xy = - 19 \\ \text{xy}(x + y) = - 20\ \ \\ \end{matrix}\text{\ \ \ } \right.\ \]

\[Пусть\ x + y = t;\ \ \ xy = c:\]

\[\left\{ \begin{matrix} t + c = - 19 \\ ct = - 20\ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ } \right.\ \ \]

\[\left\{ \begin{matrix} t = - 19 - c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ c( - 19 - c) + 20 = 0 \\ \end{matrix}\text{\ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} t = - 19 - c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - c^{2} - 19c + 20 = 0 \\ \end{matrix} \right.\ \]

\[c_{1} + c_{2} = - 19,\ \ c_{1}c_{2} = - 20\]

\[c_{1} = - 20,\ \ c_{2} = 1\]

\[\left\{ \begin{matrix} c = - 20 \\ t = 1\ \ \ \ \ \ \\ \end{matrix}\ \ \ \ \ \ или\ \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} c = 1\ \ \ \ \ \ \\ t = - 20 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} xy = - 20 \\ x + y = 1 \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} xy = 1\ \ \ \ \ \ \ \ \ \ \ \ \\ x + y = - 20 \\ \end{matrix} \right.\ \]

\[t^{2} + 20t + 1 = 0\]

\[D = 400 - 4 = 396\]

\[t_{1,2} = \frac{- 20 \pm 6\sqrt{11}}{2} =\]

\[= - 10 \pm 3\sqrt{11}\]

\[x = - 4,\ \ y = 5\]

\[y = - 4,\ \ x = 5.\]

\[Ответ:( - 4;5),\ (5;\ - 4),\ \]

\[( - 10 + 3\sqrt{11};\ - 10 - 3\sqrt{11}),\ \]

\[\left( - 10 - 3\sqrt{11};\ - 10 + 3\sqrt{11} \right).\]