Bài 1:
a: \(x^2-2y^2=xy\)
=>\(x^2-xy-2y^2=0\)
=>\(x^2-2xy+xy-2y^2=0\)
=>x(x-2y)+y(x-2y)=0
=>(x-2y)(x+y)=0
mà x+y>0(x>0 và y>0)
nên x-2y=0
=>x=2y
\(B=\dfrac{3x-y}{x+y}=\dfrac{3\cdot2y-y}{2y+y}=\dfrac{5}{3}\)
b: \(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)
=>\(3\left(x^2+y^2\right)=10xy\)
=>\(3x^2-9xy-xy+3y^2=0\)
=>\(3x\left(x-3y\right)-y\cdot\left(x-3y\right)=0\)
=>(x-3y)(3x-y)=0
=>\(\left[{}\begin{matrix}x=3y\\x=\dfrac{y}{3}\end{matrix}\right.\)
0<x<y nên x không thể bằng 3y
=>x=y/3
\(A=\dfrac{x-y}{x+y}=\dfrac{\dfrac{y}{3}-y}{\dfrac{y}{3}+y}=\dfrac{-2}{3}:\dfrac{4}{3}=\dfrac{-2}{4}=-\dfrac{1}{2}\)
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