正整數指數定律【20分】

👨🏻‍🏫 學術台數學

[匿名]

2009-11-06 5:14 am

(-3h^2k^4)^2(24h^7k^9)
我吾識點計,吾該教我=]]]
要步驟喔=]]

更新1:

(-3h^2k^4)^3÷(24h^7k^9) 我吾識點計,吾該教我=]]] 要步驟喔=]]

更新2:

呢個先喖架↑

更新3:

呢個先啱架↑

回答 (2)

用戶9112

2009-11-06 5:31 am

✔ 最佳答案

(-3h^2k^4)^3 (24h^7k^9)
= [(-3)^3h^(2*3)k^(4*3)] (24h^7k^9)
= (-27h^6k^12) (24h^7k^9)
= (-27/24)h^(6-7)k^(12-9)
= (-9/8)h^(-1)k^3
= -(9k^3)/(8h)

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Bill

2009-11-06 5:34 am

(-3h^2k^4)^3÷(24h^7k^9)
=-3^3h^6k^12÷24h^7k^9
=-27k^12-9÷24h^7-6
=-27k^3÷24h
(上下約簡佢都得)

2009-11-05 21:35:23 補充：
&divide即係除號

2009-11-05 21:36:49 補充：
=9k^3÷8h
樓上岩,我唔記得÷3

參考: own

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收錄日期: 2021-04-23 23:21:07
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