求代数式x^4-y^4/x^3+x^2y+xy^2+y^3的值,其中x=2002,y=2001

ww风56562022-10-04 11:39:541条回答

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伤往事 共回答了22个问题 | 采纳率86.4%
请楼主以后千万不要省掉两个关键的括号,浪费了我15分钟的时间.
(x^4-y^4)/(x^3+x^2y+xy^2+y^3)
分子等于(x^2+y^2)(x+y)(x-y)
分母等于(x+y)^3-2(x^2y+xy^2)
=(x+y)^3-2xy(x+y)
分子分母同时约x+y,得出原式
=(x^2+y^2)(x-y) / [(x+y)^2-2xy]
=(x^2+y^2)(x-y) / (x^2+y^2)
=x-y
=2002-2001
=1
1年前

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题有错吧.因该是+2y^4吧
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X^4-Y^4
=(x^2+y^2)(x^2-y^2)
=(x^2+y^2)(x+y)(x-y)
X^3+X^2Y+XY^2+Y^3
=x^2(x+y)+y^2(x+y)
=(x^2+y^2)(x+y)
X^4-Y^4/X^3+X^2Y+XY^2+Y^3
=x-y
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