Ta có: \(x^2-2y^2=xy\)

\(\Leftrightarrow x^2-xy-2y^2=0\)

\(\Leftrightarrow x^2-2xy+xy-2y^2=0\)

\(\Leftrightarrow x\left(x-2y\right)+y\left(x-2y\right)=0\)

\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)

Vì \(x+y\ne0\) nên x-2y=0

hay x=2y

Thay x=2y vào biểu thức \(A=\dfrac{x-y}{x+y}\), ta được: 

\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)

Vậy: \(A=\dfrac{1}{3}\)

chs bb ak

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

\(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\Leftrightarrow x^2+xy-2xy-2y^2=0\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)

Mà \(x+y\ne0\Rightarrow x-2y=0\Rightarrow x=2y\)

\(\Rightarrow A=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)

x²  - 2y² = xy <=> x² - xy - 2y² = 0 <=> x² + xy - 2xy - 2y² = 0 <=> x (  x + y ) - 2y 

( x + y ) = 0 <=> ( x - 2y ) ( x + y ) = 0

mà x + y \(\ne\) 0 => x - 2y = 0 => x = 2y

=> A = \(\frac{2y-y}{2y+y}\) = \(\frac{y}{3y}\) = \(\frac{1}{3}\)

Ta có :\(x^2-2y^2=xy\)

\(x^2-xy-2y^2=0\)

\(x^2+2xy+y^2-3xy-3y^2=0\)

\(\left(x+y\right)^2-3y\times\left(x+y\right)=0\)

\(\left(x+y\right)\left(x+y-3y\right)=0\)

\(\Rightarrow\begin{cases}x-2y=0\\x+y=0\end{cases}\)

Vậy \(\frac{x+y}{x-y}=\frac{0}{x-y}=0\)

\(x^2-2y^2=xy\Rightarrow x^2-y^2=xy+y^2\)

\(\left(x-y\right)\left(x+y\right)=y\left(x+y\right)\)

\(\Rightarrow x-y=y\)

\(x=2y\)

Thay \(x=2y\)

\(\frac{x+y}{x-y}=\frac{2y+y}{2y-y}=\frac{3y}{y}=3\)

3 đó các bạn

Từ đề bài \(\Rightarrow\)\(x^2-2y^2-xy=0\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)

Mà \(x+y\ne0\Rightarrow x-2y=0\Rightarrow x=2y\)

\(\Rightarrow P=\frac{2y-y}{2y+y}=\frac{1}{3}\)

Vì \(x^2-2y^2=xy\) 

\(\Leftrightarrow x^2-xy-y^2=0\)

\(\Leftrightarrow\left(x-y\right)^2-y\left(x+y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)

Theo đề bài thì có : 

\(x+y\ne0\)

\(\Rightarrow x-2y=0\)

\(\Leftrightarrow x=2y\)

Từ đó ta lại có :

\(P=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)

Vậy .......

ta có 

          x²-2y²=xy

<=>  x² -xy -2y² =0

<=> (x-2y)(x+y)=0

=>\(\orbr{\begin{cases}x=2y\\x+y=0\left(loại\right)\end{cases}}\)

nếu x=2y thì P=1/3