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计算分式:[(4a^2-4a)/(a^2-2a+1)]/[(a+1)/(a-1)-(a-1)/(a+1)]*[1/(3a-3)]谢谢了,
人气:111 ℃ 时间:2020-05-19 14:21:01
解答
[(4a^2-4a)/(a^2-2a+1)]/[(a+1)/(a-1)-(a-1)/(a+1)]*[1/(3a-3)] =[4a(a-1)/(a-1)^2]/[4a/(a+1)(a-1)]*[1/3(a-1)] =[4a/(a-1)]/[4a/3(a+1)(a-1)^2] =3(a+1)(a-1)
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