(a^4/3-8a^1/3b)/(4b^2/3+2三次根号ab+a^2/3)/(1-2三次根号b/a)*三次根号a

bessie_go2022-10-04 11:39:541条回答

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晨与橙愿共回答了18个问题 | 采纳率94.4%: (a^4/3-8³√ab)/(4b^2/3+2³√ab+a^2/3)/(1-2³√b/a)*³√a
=³√a(a-8b)/(4b^2/3+2³√ab+a^2/3)/(1-2³√b/a)*³√a
=³√a(³√a-2³√b)(a^2/3+2³√ab+4b^2/3)/(4b^2/3+2³√ab+a^2/3)/(1-2³√b/a)*³√a
=³√a(³√a-2³√b)/(³√a-2³√b)
=³√a; 1年前

化简（a^4/3-8a^1/3b）/（4b^2/3+2根号下3次方ab+a^2/3）÷（a^-2/3-2根号下三次方b/ 化简（a^4/3-8a^1/3b）/（4b^2/3+2根号下3次方ab+a^2/3）÷（a^-2/3-2根号下三次方b/a）×根号下a·根号下三次方a²/根号下五次方根号a·根号下三次方a 凉月满天20061年前1: 男子汉36318 共回答了23个问题 | 采纳率87%

(a^4/3-8³√ab)/(4b^2/3+2³√ab+a^2/3)/(1-2³√b/a)*³√a
=³√a(a-8b)/(4b^2/3+2³√ab+a^2/3)/(1-2³√b/a)*³√a
=³√a(³√a-2³√b)(a^2/3+2³√ab+4b^2/3)/(4b^2/3+2³√ab+a^2/3)/(1-2³√b/a)*³√a
=³√a(³√a-2³√b)/(³√a-2³√b)
=³√a

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(a^4/3-8a^1/3b)/(a^2/3+2(ab)^1/3+4b^2/3)除以(1-2(b/a)^1/3)×a^1 (a^4/3-8a^1/3b)/(a^2/3+2(ab)^1/3+4b^2/3)除以(1-2(b/a)^1/3)×a^1/3 化简 飞龙在南1年前1: BILLCHAN 共回答了15个问题 | 采纳率80%

(a^4/3-8a^1/3b)/a^2/3+2(ab)^1/3+4b^2/3/(1-2(b/a)^1/3)*a^1/3
=[a^1/3(a-8b)]/[a^2/3+2(ab)^1/3+4b^2/3]÷[1-(2b^1/3/a^1/3)]*a^1/3
=[a^1/3(a^1/3-2b^1/3)(a^2/3+2a^1/3b^1/3+4b^2/3)]/[a^2/3+2(ab)^1/3+4b^2/3]÷[(a^1/3-2b^1/3)/a^1/3]*a^1/3
=[a^1/3(a^1/3-2b^1/3)*[a^1/3/(a^1/3-2b^1/3)]*a^1/3
=a^(1/3+1/3+1/3)
=a

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