Ta có: \(A=\dfrac{x^2+y^2}{x-y}=\dfrac{x^2-2xy+y^2+2xy}{x-y}=\dfrac{\left(x-y\right)^2+2}{x-y}=\dfrac{\left(x-y\right)^2}{x-y}+\dfrac{2}{x-y}=\left(x-y\right)+\dfrac{2}{x-y}\)
Vì \(x>y\Rightarrow x-y>0\)
Áp dụng bđt cô si cho 2 số dương (x-y) và \(\dfrac{2}{x-y}\) có:
\(\left(x-y\right)+\dfrac{2}{x-y}\ge2\sqrt{\dfrac{2\left(x-y\right)}{x-y}}=2\sqrt{2}\)
Vậy \(MIN_A=2\sqrt{2}\)