Решите систему уравнений: 3/(x+y)+1/(y-x)=6/(x^2-y^2); (x-2)^2/(y+2)=1.

\[\left\{ \begin{matrix} \frac{3}{x + y} + \frac{1}{y - x} = \frac{6}{x^{2} - y^{2}} \\ \frac{(x - 2)^{2}}{y + 2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

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

\[4y^{2} + 4y + 1 - y - 2 = 0\]

\[4y^{2} + 3y - 1 = 0\]

\[D = 3^{2} - 4 \cdot 4 \cdot ( - 1) =\]

\[= 9 + 16 = 25;\ \ \ \ \ \sqrt{D} = 5.\]

\[y_{1} = \frac{- 3 + 5}{2 \cdot 4} = \frac{2}{8} = \frac{1}{4};\ \ \ \ \ \]

\[\text{\ \ }y_{2} = \frac{- 3 - 5}{2 \cdot 4} = \frac{- 8}{8} = - 1\]

\[x_{1} = 2 \cdot \frac{1}{4} + 3 = \frac{1}{2} + 3 = 3,5;\ \ \ \ \]

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

\[= 1\ \ (не\ подходит).\]

\[Ответ:\left( 3,5;\frac{1}{4} \right).\]