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通分两个式子(A-1)/[(A+1)平方-4],(1-A)/(2-4A+2A平方)
人气:352 ℃ 时间:2020-03-26 16:29:56
解答
(A-1)/[(A+1)平方-4]
=(A-1)/(A+1-2)(A+1+2)
=(A-1)/(A-1)(A+3)
=1/(A+3)
(1-A)/(2-4A+2A平方)
=(1-A)/2(1-2A+A^2)
=(1-A)/2(1-A)^2
=1/2(1-A)
=-1/2(A-1)
通分分母为2(A-1)(A+3)
所以(A-1)/[(A+1)平方-4]=2(A-1)/2(A+3)(A-1)
(1-A)/(2-4A+2A平方)=-(A+3)/2(A+3)(A-1)
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