1.(x^2/3y^-3)^-2/(6x^-1y^3)
plz help!
答案對有分!因為我本書有答案,但我唔識點列式!
請問一些零指數及負指數既問題
2006-10-25 4:54 am
回答 (1)
2006-10-25 6:38 am
✔ 最佳答案
(x^2/3y^-3)^-2/(6x^-1y^3)= [x^(-4)/3y^6]*[6xy^(-3)]
= (6/3)[x^(-4)x][y^(-6)y^(-3)]
= 2 x^(-3) y^(-9)
or = 2/(x^3y^9)
收錄日期: 2021-04-12 22:59:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061024000051KK03936