老师出了一道题,化简求值[(1/y)-(1/x)][(2y^2)/(x^2-y^2)-(x-y)/2(x+y)+(x+y)/2(x-y)]

原式=[(x-y)/xy][4y^2/2(x-y)(x+y)-(x-y)^2/2(x-y)(x+y)+(x+y)^2/2(x-y)(x+y)]
=[(x-y)/xy]{[4y^2-(x-y)^2+(x+y)^2]/2(x-y)(x+y)}
=[(x-y)/xy][4y^2+4xy/2(x-y)(x+y)]
=[(x-y)/xy][4y(x+y)/2(x-y)(x+y)]
=[(x-y)/xy][2y/(x-y)]
=2/x
=-3
如果有哪里不懂可以再问,乐意帮你解答,