解(1)
原式=2(x-y)/(x²-y²)+4y/(x²-y²)
=(2x-2y+4y)/(x²-y²)
=(2x+2y)/(x²-y²)
=2(x+y)/[(x+y)(x-y)]
=2/(x-y)
(2)：
原式=x²/(x+y)-(x-y)
=x²/(x+y)-(x-y)(x+y)/(x+y)
=x²/(x+y)-(x²-y²)/(x+y)
=(x²-x²+y²)/(x+y)
=y²/(x+y)