解分式方程:【(x-y分之1)-(x+y分之1)】·【(y+1)²-(y-1)²】分之【x²-2xy+y²】

问题描述:

解分式方程:【(x-y分之1)-(x+y分之1)】·【(y+1)²-(y-1)²】分之【x²-2xy+y²】
【(x-y分之1)-(x+y分之1)】×【(y+1)²-(y-1)²】分之【x²-2xy+y²】,上面打错了

[(x-y)分之1-(x+y)分之1]·[(y+1)²-(y-1)²]分之(x²-2xy+y²)=[(x+y)(x-y)]分之[(x+y)-(x-y)]·[(y+1-y+1)(y+1+y-1)]分之(x-y)²=[(x+y)(x-y)]分之(2y)·(2y)分之(x-y)²=(x+y)分之(x-y)...