Ta có \(x^2-2y^2=xy\)
\(\Leftrightarrow x^2-y^2-y^2-xy=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-y-y\right)=0\)
\(\Leftrightarrow\begin{matrix}x+y=0&\left(1\right)&\\x-2y=0&\left(2\right)&\end{matrix}\)
vì nếu x+y=0 thì \(\frac{x-y}{x+y}\) vô nghĩa nên ta loại (1)
Do đó x-2y=0 <=>x=2y
Vậy \(\frac{x-y}{x+y}\) =\(\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)