求步驟
唔該幫幫手!!
9(4^n)-3(2^n+1)/5(2^n-1)
^後的是index
這題的答案是(18(2^n)-12)/5
但我怎都計不到
請各位幫幫手
謝謝!!!
f.3maths 題
2008-08-05 11:24 pm
回答 (2)
2008-08-05 11:49 pm
✔ 最佳答案
As follows~圖片參考：http://i248.photobucket.com/albums/gg191/ncy_0916/maths30.jpg?t=1217922562
2008-08-05 15:50:32 補充：
根據指數定律，
a^m X a^n = a^(m + n)
a^m ÷ a^n = a^(m - n)
(a^m)^n = a^(mn)
2008-08-05 15:56:43 補充：
因為2^(2n) = 2^(n + n) = (2^n)(2^n)
指數定律：a^m X a^n = a^(m + n)
2008-08-05 11:53 pm
To get the answer stated, the expression should be written as:
{9(4^n) - 3[2^(n+1)]}/5[2^(n-1)].
For easy reading, let 2^n = x.
Therefore, 9(4^n) = 9x^2.
3[2^(n+1)] = 3[2x] = 6x.
2^(n-1) = x/2. Therefore, the expression becomes
(9x^2 - 6x)/[5(x/2)] = 2x(9x - 6)/5x = (18x - 12)/5 = [18(2^n) - 12]/5.
{9(4^n) - 3[2^(n+1)]}/5[2^(n-1)].
For easy reading, let 2^n = x.
Therefore, 9(4^n) = 9x^2.
3[2^(n+1)] = 3[2x] = 6x.
2^(n-1) = x/2. Therefore, the expression becomes
(9x^2 - 6x)/[5(x/2)] = 2x(9x - 6)/5x = (18x - 12)/5 = [18(2^n) - 12]/5.
收錄日期: 2021-04-25 22:35:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080805000051KK01753