化简(1/(x-y)-1/(x+y))/(2y/(x^2-2xy+y^2))

化简(1/(x-y)-1/(x+y))/(2y/(x^2-2xy+y^2))
要具体的过程
jytjun123 1年前 已收到2个回答 举报

songsons51771 幼苗

共回答了12个问题采纳率:83.3% 举报

(1/(x-y)-1/(x+y))/(2y/(x^2-2xy+y^2))
先对分子通分,公分母是(x-y)(x+y)
=([(x+y)-(x-y)]/[(x-y)(x+y)])÷[2y/(x-y)²]
=(2y/[(x-y)(x+y)])÷[2y/(x-y)²]
=(x-y)²/[(x-y)(x+y)]
约分,上下同时约去公因子(x-y)
=(x-y)/(x+y)
公式:x²-2xy+y²=(x+y)²

1年前

2

飘零逝水 幼苗

共回答了57个问题 举报

(1/(x-y)-1/(x+y))/(2y/(x^2-2xy+y^2))
=(1/(x-y)-1/(x+y))/(2y/(x-y)^2)
=(2y/(x-y)(x+y))/(2y/(x-y)^2)
=(x-y)/(x+y)

1年前

2
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