Ta có: \(x^2-2y^2=xy\)
\(\Leftrightarrow x^2-y^2-y^2-xy=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)
Mà \(x+y\ne0\)
\(\Rightarrow x-2y=0\)
\(\Rightarrow x=2y\)
\(\Rightarrow P=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)
Đặc P ta có
P= x2 - 2y2 =xy
<=> x2 - y2 - y2 -xy =0
=> (x-1) (x+y) -y (x+y) -1
=> (x+y_(x-2y)=0
Vậy
x+y #0
=> x- 2y =0
=>x=2y
=>P=2y -y trên 2y + y =y trên 3y =1/3