(1)
(5a^1234567890)÷[(5a)^(-8765432110)]
(2)
(x^-40)÷[x^(-1/40)]
數學知識交流---知識送大禮之指數定律
2011-09-16 3:18 am
回答 (2)
2011-09-16 3:53 am
✔ 最佳答案
(1) (5a^1234567890) / [(5a)^(-8765432110)]= 5a^1234567890 / [5^(-8765432110) * a^(-8765432110)]
= 5^8765432111 * a^10000000000
(2) (x^-40) / [x^(-1/40)]
= x^[-40 - (-1/40)]
= x^(-1599/40)
= 1/x^(1599/40)
參考: Knowledge is power.
2011-09-16 4:17 am
(1)
(5a^1234567890)÷[(5a)^(-8765432110)]
= (5a) ^ (1234567890 - (-8765432110))
= (5a) ^ (10^10)
(2) (x^-40) / [x^ (-1/40)]
= x^ (-40 - (-1/40))
= 1/ (x^ 1599/40)
(5a^1234567890)÷[(5a)^(-8765432110)]
= (5a) ^ (1234567890 - (-8765432110))
= (5a) ^ (10^10)
(2) (x^-40) / [x^ (-1/40)]
= x^ (-40 - (-1/40))
= 1/ (x^ 1599/40)
收錄日期: 2021-04-27 16:36:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110915000051KK00587