Решите систему уравнений: x^2y^2-xy=12; x+y=2.

\[\left\{ \begin{matrix} x²y² - xy = 12\ (*) \\ x + y = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(*)\ x^{2}y^{2} - xy - 12 = 0\]

\[D = 1 + 48 = 49\]

\[xy_{1,2} = \frac{1 \pm 7}{2} = 4;\ - 3.\]

\[1)\ \left\{ \begin{matrix} xy = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x + y = 2 \Longrightarrow x = 2 - y \\ \end{matrix} \right.\ \]

\[(2 - y)y = 4\]

\[2y - y^{2} = 4\]

\[y^{2} - 2y + 4 = 0\]

\[D = 4 - 8 = 4 < 0 \rightarrow \varnothing.\]

\[2)\ \left\{ \begin{matrix} xy = - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x + y = 2 \Longrightarrow x = 2 - y \\ \end{matrix} \right.\ \]

\[(2 - y)y = - 3\]

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

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

\[По\ теореме\ Виета:\]

\[y_{1} = 3;\ \ y_{2} = - 1.\]

\[x_{1} = 2 - 3 = - 1.\]

\[x_{2} = 2 - ( - 1) = 3.\]

\[Ответ:( - 1;3);(3;\ - 1).\]