Решите систему уравнений x^2-y^2=4*(x+y); 1/(5x-4y)=1/9.

\[\left\{ \begin{matrix} x^{2} - y^{2} = 4(x + y) \\ \frac{1}{5x - 4y} = \frac{1}{9}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} (x + y)(x - y - 4) = 0 \\ 5x - 4y = 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\left\{ \begin{matrix} x + y = 0\ \ \ \ \ \\ 5x - 4y = 9 \\ \end{matrix} \right.\ \ \]

\[x = - y:\]

\[- 5y - 4y = - 9\]

\[- 9y = - 9\]

\[y = 1.\]

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

\[2)\ \left\{ \begin{matrix} x - y = 4\ \ \ \ \ \\ 5x - 4y = 9 \\ \end{matrix} \right.\ \]

\[x = 4 + y\]

\[5(4 + y) - 4y = 9\]

\[20 + 5y - 4y = 9\]

\[y = - 11;\]

\[x = 4 - 11 = - 7.\]

\[Ответ:( - 1;1);( - 7;\ - 11).\]