數學指數問題exponent..
回答 (1)
2008-09-27 12:46 am
✔ 最佳答案
[16^(n + 2)]^(1/3) = [2^(4n + 8)]^(1/3) = 2^[(4n + 8)/3] = A[32^( n -1)]^(1/5) = [2^(5n -5)]^(1/5) = 2^(n -1). = B.
So A/B = [2^(4n + 8)/3]/[2^(n -1)] = 2^[(4n + 8)/3 - (n -1)]
= 2^[( 4n + 8 - 3n + 3)/3] = 2^[(n + 11)/3]
收錄日期: 2021-04-25 22:35:14
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https://hk.answers.yahoo.com/question/index?qid=20080926000051KK00937