chunge
幼苗
共回答了23个问题采纳率：91.3% 举报
原式=(把前两个局部通分)[(x+2y)(x-2y)+4y^2]/(x-2y)+4x^2y/(2y+x)(2y-x)
=(前两个局部计算) x^2/(x-2y)-4x^2y/(x+2y)(x-2y):[注意最后一项的分母变成了：x-2y,所以前面的符号变成了：-]
=(在与最后的分式通分) [x^2(x+2y)-4x^2y]/(x+2y)(x-2y)
=(加减计算)(x^3-2x^2y)/(x+2y)(x-2y)
=x^2(x-2y)/(x+2y)x-2y)
=(约分化简) x^2/(x+2y)
1年前
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