Решите систему уравнений: xy=-12; x^2+y^2=25.

\[(1) + (2):\ \ \ \ \ \ (x + y)^{2} = 1\]

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

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

\[D = ( - 1)^{2} - 4 \cdot 1 \cdot ( - 12) =\]

\[= 1 + 48 = 49;\ \ \ \ \sqrt{D} = 7.\]

\[y_{1} = \frac{1 + 7}{2} = \frac{8}{2} = 4;\ \ \ \ \ \]

\[\text{\ \ }y_{2} = \frac{1 - 7}{2} = \frac{- 6}{2} = - 3\]

\[x_{1} = 1 - 4 = - 3;\ \ \ \ \ \ \ \ \]

\[\text{\ \ \ \ }x_{2} = 1 - ( - 3) = 1 + 3 = 4.\]

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

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

\[D = 1^{2} - 4 \cdot 1 \cdot ( - 12) =\]

\[= 1 + 48 = 49;\ \ \ \sqrt{D} = 7.\]

\[y_{1} = \frac{- 1 + 7}{2} = \frac{6}{2} = 3;\ \ \ \ \ \ \]

\[\ y_{2} = \frac{- 1 - 7}{2} = \frac{- 8}{2} = - 4\]

\[x_{1} = - 3 - 1 = - 4;\ \ \ \ \ \ \ \ \ \]

\[\text{\ \ \ }x_{2} = - ( - 4) - 1 = 4 - 1 = 3.\]

\[Ответ:( - 3;4);\ \ \ (4;\ - 3),;\ \]

\[\ ( - 4;3);\ \ \ (3;\ - 4).\]