Question

$\left \{ {{\frac{x+y}{x-y}- \frac{2(x-y)}{x+y}=1 } \atop {x^2-5xy+2y^2=4}} \right.$
Пожалуйста, помогите решить систему уравнения.

Answer 1

$\begin{cases}\frac{x+y}{x-y}- \frac{2(x-y)}{x+y}=1\\x^2-5xy+2y^2=4\end{cases}$

$\begin{cases}\frac{(x+y)^2-2(x-y)^2}{(x-y)(x+y)}=1\\x^2-5xy+2y^2=4\end{cases}$

$\begin{cases}\frac{x^2+2xy+y^2-2x^2+4xy-2y^2}{x^2-y^2}=1\ /\cdot(x^2-y^2)\\x^2-5xy+2y^2=4\end{cases}$

$\begin{cases}-x^2+6xy-y^2=x^2-y^2\\x^2-5xy+2y^2=4\end{cases}$

$\begin{cases}-x^2+6xy-y^2-x^2+y^2=0\\x^2-5xy+2y^2=4\end{cases}$

$\begin{cases}-2x^2+6xy=0\ /:(-2)\\x^2-5xy+2y^2=4\end{cases}$

$\begin{cases}x^2-3xy=0\\x^2-5xy+2y^2=4\end{cases}$

$\begin{cases}x(x-3y)=0\\x^2-5xy+2y^2=4\end{cases}$

$\begin{cases}x=0\\x^2-5xy+2y^2=4\end{cases}\ \ \ \ \ \ \ \ \ \begin{cases}x=3y\\x^2-5xy+2y^2=4\end{cases}$

$\begin{cases}x=0\\0^2-5 \cdot 0 \cdot y+2y^2=4\end{cases}\ \ \ \ \ \ \begin{cases}x=3y\\(3y)^2-5 \cdot 3y \cdot y+2y^2=4\end{cases}$

$\begin{cases}x=0\\2y^2=4\ /:2\end{cases}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{cases}x=3y\\9y^2-15y^2+2y^2=4\end{cases}$

$\begin{cases}x=0\\y^2=2\end{cases}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{cases}x=3y\\-4y^2=4\ /:(-4)\end{cases}$

$\begin{cases}x=0\\y^2-2=0\end{cases}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{cases}x=3y\\y^2=-1\end{cases}$

$\begin{cases}x=0\\(y-\sqrt2)(y+\sqrt2)=0\end{cases}\ \ \ \ \ \ \ \ \begin{cases}x=3y\\y \in\emptyset\end{cases}$

$\begin{cases}x=0\\y=\sqrt2\end{cases}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{cases}x=0\\y=-\sqrt2\end{cases}$

Ответ: (0,√2),(0,-√2)