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| 2017/06/21 07:29:36瀏覽44｜回應0｜推薦0 | |
| 標題: 中4數學…指數…PLEASE 發問: (4^n+3)+(32^n+1) / (2^2n+1)+(2^5n) 步驟please... 最佳解答: [(4^(n+3)+32^(n+1)]/[(2^(2n+1)+2^(5n)] =[2^(2n+6)+2^(5n+5)]/[2^(2n+1)+2^(5n)] =2^5[2^(2n+1)+2^(5n)]/[2^(2n+1)+2^(5n)] =32 其他解答: (4^n+3)+(32^n+1) / (2^2n+1)+(2^5n) =36^n+4/2^10n+1 =18^9n+5
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