How can I solve (27a^-3b^6c^3)^1/3?
回答 (4)
2010-07-06 5:27 pm
✔ 最佳答案
.. (27 a^-3 b^6 c^3)^1/3= (27)^1/3 (a^-3)^1/3 (b^6)^1/3 (c^3)^1/3
= 3 (a^-1) (b^2) (c^1)
= 3b²c/a
2010-07-06 5:21 pm
3b^2c/a
2010-07-06 5:25 pm
(27a^-3b^6c^3)^1/3
=27^(1/3) a^-3*(1/3) b^6*(1/3) c^3*(1/3) using (a^x)^y = a^(xy) and (ab)^x=a^x b^x
=3a^(-1) b^2 c
3 b^2 c
= -----------
a
=27^(1/3) a^-3*(1/3) b^6*(1/3) c^3*(1/3) using (a^x)^y = a^(xy) and (ab)^x=a^x b^x
=3a^(-1) b^2 c
3 b^2 c
= -----------
a
參考: me
收錄日期: 2021-05-03 14:25:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100706091730AAKyhD4