請詳細步驟教我計下條 :
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Solve (n+3)(n+2)!=120 x n!
急 ! F5 數1條,solve ! 的計法
2012-10-28 7:20 am
回答 (2)
2012-10-28 7:46 am
✔ 最佳答案
(n+3)(n+2)!=120 x n! (n+3)(n+2)(n+1)n!=120 x n!
(n+3)(n+2)(n+1)=120
n^3 + 6n^2 +11n - 114 =0
(n-3)(n^2 + 9n + 38) = 0
n = 3
2012-10-28 7:44 am
(n+3)(n+2)!=120 x n!
(n+3)(n+2)(n+1)n!=120xn!
(n+3)(n+2)(n+1)=120
n=3
(n^2+5n+6)(n+1)=120
n^3+5n^2+6n+n^2+5n+6=120
n^3+6n^2+11n-114=0
(n-3)(n^2+9n+38)=0
Because n^2+9n+38 has no real root
Therefore, n=3
(n+3)(n+2)(n+1)n!=120xn!
(n+3)(n+2)(n+1)=120
n=3
(n^2+5n+6)(n+1)=120
n^3+5n^2+6n+n^2+5n+6=120
n^3+6n^2+11n-114=0
(n-3)(n^2+9n+38)=0
Because n^2+9n+38 has no real root
Therefore, n=3
參考: myself
收錄日期: 2021-04-13 19:05:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121027000051KK00695