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求代数式x³+x²y+xy²+y³分之x的四次方-y的四次方的值,期中x=2002,y=2001
人气:431 ℃ 时间:2019-08-18 03:32:25
解答
原式=(x²+y²)(x²-y²)/[x(x²+y²)+y(x²+y²)]
=(x²+y²)(x+y)(x-y)/[(x+y)(x²+y²)]
=x-y
=1
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