xiaoqin612
幼苗
共回答了11个问题采纳率:90.9% 举报
(x+y/x-y+4xy/y²-x²)/x²-4y²/x²+3xy+2y²
分子A=(x+y)/(x-y)+(4xy)/(y²-x²)
=[(x+y)²-4xy]/(x²-y²)
=(x-y)²/(x²-y²)
=(x-y)/(x+y)
分母B=(x²-4y²)/(x²+3xy+2y²)
=[(x-2y)(x+2y)]/[(x+y)(x+2y)]
=(x-2y)/(x+y)
A/B=[(x-y)/(x+y)]/[(x-2y)/(x+y)]
=(x+y)/(x-2y)
当2x²+3xy-5y²=0时,有(2x+5y)(x-y)=0,即2x+5y=0或x-y=0,显然后者不合题意,故有
x=-5/2y,代入(x+y)/(x-2y)可得原式的值=5/12
1年前
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