(x^2 - 2xy + y^2) / ((x^2 - y^2) (x - y))
将分子进行因式分解，得到：
x^2 - 2xy + y^2 = (x - y)^2
将分母进行因式分解，得到：
(x^2 - y^2) (x - y) = (x + y) (x - y) (x - y) = (x - y)^2 (x + y)
将分子和分母的因式化简代入原式，得到：
(x - y)^2 / ((x - y)^2 (x + y)) = 1 / (x + y)
因此，原式的值为 1 / (x + y)。
15.(2022赣州崇义期中)先化简,再求值, (x^2-2xy+y^2)\/(x^2-y^2)(x?
(x^2 - 2xy + y^2) \/ ((x^2 - y^2) (x - y))将分子进行因式分解，得到：x^2 - 2xy + y^2 = (x - y)^2 将分母进行因式分解，得到：(x^2 - y^2) (x - y) = (x + y) (x - y) (x - y) = (x - y)^2 (x + y)将分子和分母的因式化简代入原式，得到...
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先化简,再求值:(x的平方-2xy+y的平方)-(x的平方+2xy+y的平方),其中x=...
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我的 先化简,再求值:[(x^2-y^2)\/(x^2+2xy+y^2)]\/[(x-y)\/x^2+xy],其中,x=2012,y=2013 1个回答 #热议# 有哪些跨界“双奥”的运动员? wsp竹木 2014-04-24 · TA获得超过288个赞 知道小有建树答主 回答量:59 采纳率:0% 帮助的人:65.1万 我也去答题访问个人页 关注 展开全...
先化简再求值x-(x\/y)^2+y^2-2xy除以
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...\\(x^2+2xy+y^2)÷x\\(x-y)×(x+y)\\(x^2-2xy+y^2),已知x=1\\3,y=...
(x^2y-x^3)\/(x^2+2xy+y^2)÷x\/(x-y)×(x+y)\/(x^2-2xy+y^2)=x^2(y-x)\/(x+y)^2÷x\/(x-y)×(x+y)\/(x-y)^2 =x^2(y-x)\/(x+y)^2×(x-y)\/x×(x+y)\/(x-y)^2 =x(y-x)\/(x+y)×1\/(x-y)=-x(x-y)\/(x+y)×1\/(x-y)=-x\/(x+y)=(-1\/...
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(x^2-2xy)^2+2y^2(x^2-2xy)+y^4
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