4^(2d-2)/64^d-2
點計?
數學問題一條
2007-07-23 2:46 am
回答 (2)
2007-07-23 3:16 am
✔ 最佳答案
4^(2d-2) / 64^(d-2)= 4^(2d-2) / 4^(3d-6)
= 4^[(2d-2)-(3d-6)]
= 4^(4-d)/
2007-07-23 3:45 am
4^(2d-2)/64^d-2
=4^[2(d-1)]/4^[3(d-2)]
=4^[2(d-1)-3(d-2)]
=4^[2d-2-3d+6]
=4^(-d+4)
希望可以幫到你
=4^[2(d-1)]/4^[3(d-2)]
=4^[2(d-1)-3(d-2)]
=4^[2d-2-3d+6]
=4^(-d+4)
希望可以幫到你
參考: myself
收錄日期: 2021-04-19 22:23:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070722000051KK03162