问题标题:
【一道高一数学指数式计算题,求方法!计算((a^4/3-8(ab)^1/3)/(a^2/3+2(ab)^1/3+4a^4/3))/(1-2(b/a)^1/3)=_____】
问题描述:

一道高一数学指数式计算题,求方法!

计算((a^4/3-8(ab)^1/3)/(a^2/3+2(ab)^1/3+4a^4/3))/(1-2(b/a)^1/3)=_____

廖清裕回答:
  ((a^4/3-8(ab)^1/3)/(a^2/3+2(ab)^1/3+4a^4/3))/(1-2(b/a)^1/3)   =a^(1/3)*(a-8b^(1/3))/(a^(1/3)*(a^(1/3)+2b^(1/3)+4a))/(1-2(b/a)^(1/3))   =(a-8b^(1/3))/(a^(1/3)+2b^(1/3)+4a)/(1-2(b/a)^(1/3))
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