Question

[Trương Lan Anh]

tính giá trị A=\(\frac{x-y}{x+y}\) biết \(x^2-2y^2=xy\left(y\ne0,x+y\ne0\right)\)

Answer 1

[ST]

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

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

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

Answer 2

[huy chiến]

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