Question 34474
(x-y)/(x^2-y^2) + (x+y)/(x^2-y^2) - 2x/(y^2-x^2)
= (x-y)/(x^2-y^2) + (x+y)/(x^2-y^2) - 2x/(-1)(x^2-y^2)
= (x-y)/(x^2-y^2) + (x+y)/(x^2-y^2) + 2x/(x^2-y^2)
= [1/(x^2-y^2)][(x-y)+(x+y)+2x] = [1/(x^2-y^2)][4x]
= 4x/(x^2-y^2)