$ĐKXĐ : x \neq y $
Ta có :
$A=\dfrac{1}{x-y}+\dfrac{3xy}{x^3-y^3}+\dfrac{x-y}{x^2+xy+y^2} $
$=\dfrac{x^2+xy+y^2+3xy+(x-y).(x-y)}{(x-y).(x^2+xy+y^2)} $
$=\dfrac{x^2+xy+y^2+3xy+x^2+y^2-2xy}{(x-y).(x^2+xy+y^2)}$
$=\dfrac{2x^2+2xy+2y^2}{(x-y).(x^2+xy+y^2)} $
$=\dfrac{2}{x-y} $