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

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

你这题估计你打错了,最后一个应该是y^3

(x^4-y^4)/(x^3+x^2y+xy^2+y^3)
=(x+y)(x-y)(x^2+y^2)/[(x+y)(x^2y^2)]
=x-y
=2002-2001
=1

x-y=1
(x^4-y^4)/(x^3+x^2y+xy^2+y^3)
=(x^2+y^2)(x^2-y^2)/[x^2(x+y)+y^2(x+y)]
=(x^2+y^2)(x+y)(x-y)/(x+y)(x^2+y^2)
=x-y
=1

(x^4-y^4)/(x^3+x^2y+xy^2+y^3)
=(x^2-y^2)(x^2+y^2)/(x^2+y^2)(x+y)
=(x-y)
=2002-2001
=1