计算[-(1/2a²x)²+(1/3ax²×ax)]÷(1/6ax)
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计算[-(1/2a²x)²+(1/3ax²×ax)]÷(1/6ax)
[-(1/2a²x)²+(1/3ax²×ax)]÷(1/6ax)
=(-1/4a^4·x^2+1/3a^2·x^3)÷(1/6ax)
=-3/2a^3·x+2a·x^2
再问: [-(1/2a²x)²+(1/3ax²×ax)]÷(-1/6ax) 题目弄错了,麻烦在看下
再答: [-(1/2a²x)²+(1/3ax²×ax)]÷(-1/6ax) =(-1/4a^4·x^2+1/3a^2·x^3)÷(-1/6ax) =3/2a^3·x-2a·x^2
=(-1/4a^4·x^2+1/3a^2·x^3)÷(1/6ax)
=-3/2a^3·x+2a·x^2
再问: [-(1/2a²x)²+(1/3ax²×ax)]÷(-1/6ax) 题目弄错了,麻烦在看下
再答: [-(1/2a²x)²+(1/3ax²×ax)]÷(-1/6ax) =(-1/4a^4·x^2+1/3a^2·x^3)÷(-1/6ax) =3/2a^3·x-2a·x^2
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