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(1/a+3)-(6/9-a的平方)=?
人气:308 ℃ 时间:2020-06-20 17:35:24
解答
1/(a+3)-6/(9-a²)
=1/(a+3)+6/(a²-9)
=1/(a+3)+6/[(a+3)(a-3)]
=(a-3)/[(a+3)(a-3)]+6/[(a+3)(a-3)]
=(a+3)/[(a+3)(a-3)]
=1/(a-3)
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