Ta có: \(A=\dfrac{x-y}{x+y}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\)
\(=\dfrac{x^2-y^2}{x^2+2xy+y^2}\)
Ta có: \(x^2+2xy+y^2>x^2+y^2\forall x>y>0\)
\(\Leftrightarrow\dfrac{x^2-y^2}{x^2+2xy+y^2}< \dfrac{x^2-y^2}{x^2+y^2}\)
hay A<B