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2道分式乘除
1/(x-3)^2 × (x-3)/(x+3) × (3-x)(3+x)
(1-x)^2(3+x)^2/(x-y)^4(x-1)^2
人气:281 ℃ 时间:2020-02-05 04:52:32
解答
1/(x-3)^2 × (x-3)/(x+3) × (3-x)(3+x)
=1/[(x-3) × (x+3)] × (3-x)(3+x)
=1/[(x-3) × (x+3)] × (3-x)(x+3)
=1/(x-3) × (3-x)
=-1/(3-x)× (3-x)
=-1
(1-x)^2(3+x)^2/(x-y)^4(x-1)^2
=(x-1)^2(3+x)^2/(x-y)^4(x-1)^2
=(3+x)^2/(x-y)^4
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