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

Ответ:

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

\[Пусть\ \frac{x + y}{x - y} = t:\]

\[t - \frac{2}{t} - 1 = 0\ \ \ \]

\[t^{2} - t - 2 = 0\]

\[t_{1} + t_{2} = 1;\ \ t_{1}t_{2} = - 2,\ \ \]

\[t_{1} = - 1;\ \ t_{2} = 2\]

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

\[\left\{ \begin{matrix} x + y = y - x\ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5xy + 2y^{2} = 4 \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} 2x = 0,\ \ \ \ \ \ \ \ \ \ \ \ \ x = 0 \\ x^{2} - 5xy + 2y^{2} = 4 \\ \end{matrix} \right.\ \]

\[0 - 5 \cdot 0 + 2y^{2} = 4\ \ \]

\[y = \pm \sqrt{2}.\]

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

\[\left\{ \begin{matrix} x + y = 2x - 2y\ \ \ \ \ \\ x^{2} - 5xy + 2y^{2} = 4 \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x = 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 9y^{2} - 15y^{2} + 2y^{2} = 4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 3y\ \ \ \\ - 4y = 4 \\ \end{matrix}\text{\ \ \ \ } \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 3y\ \ \\ y^{2} = - 1 \\ \end{matrix}\ \rightarrow \ нет\ корней. \right.\ \]

\[Ответ:\left( 0;\ - \sqrt{2} \right),\ \left( 0;\ \sqrt{2} \right).\]