Question

x/(x^2+y^2) - y(x-y)^2/(x^4-y^4)

Answer 1

$\dfrac{x}{x^2+y^2} - \dfrac{y(x-y)^2}{x^4-y^4} =\dfrac{x}{x^2+y^2} - \dfrac{y(x-y)^2}{(x^2+y^2)(x^2-y^2)} =$

$=\dfrac{x}{x^2+y^2} - \dfrac{y(x-y)^2}{(x^2+y^2)(x+y)(x-y)} =$

$=\dfrac{x(x+y)}{(x^2+y^2)(x+y)} - \dfrac{y(x-y)}{(x^2+y^2)(x+y)} =$

$=\dfrac{x(x+y)-y(x-y)}{(x^2+y^2)(x+y)} =\dfrac{x^2+xy-xy+y^2}{(x^2+y^2)(x+y)} =\dfrac{x^2+y^2}{(x^2+y^2)(x+y)} =\dfrac{1}{x+y}$