化簡
k(k+1)2 + (k+1)(3k+4)
化簡 (1 question)
2007-10-30 5:47 am
回答 (5)
2007-10-30 5:51 am
✔ 最佳答案
k(k+1)2 + (k+1)(3k+4)= [ k(k+1) + (3k+4) ] (k+1)
= (k2 + k + 3k + 4) (k+1)
= (k2 + 4k + 4) (k+1)
= (k + 2)2 (k+1)
2007-11-02 4:43 am
k(k+1)^2 + (k+1)(3k+4)
=k(k^2+2k+1)+3k^2+4k+3k+4
=k^3+2k^2+k+3k^2+4k+3k+4
=k^3+5k^2+7k+4
2007-11-01 20:48:16 補充：
(k 1)^2=k^2 2k 1係因為可以變成(k 1)(k 1)再變成k．ｋ＋ｋ＋ｋ＋１ =k^2 2k 1→係由此推出___
2007-11-01 20:49:03 補充：
打錯係(k 1)^2=k^2 2k 1係因為可以變成(k 1)(k 1)再變成k．ｋ＋ｋ＋ｋ＋１=k^2 2k 1→係由此推出___
=k(k^2+2k+1)+3k^2+4k+3k+4
=k^3+2k^2+k+3k^2+4k+3k+4
=k^3+5k^2+7k+4
2007-11-01 20:48:16 補充：
(k 1)^2=k^2 2k 1係因為可以變成(k 1)(k 1)再變成k．ｋ＋ｋ＋ｋ＋１ =k^2 2k 1→係由此推出___
2007-11-01 20:49:03 補充：
打錯係(k 1)^2=k^2 2k 1係因為可以變成(k 1)(k 1)再變成k．ｋ＋ｋ＋ｋ＋１=k^2 2k 1→係由此推出___
參考: 我
2007-10-30 11:36 pm
k(k+1)2+(k+1)(3k+4)
=(k+1)[k(k+1)+(3k+4)]
=(k+1)[k^2+k+3k+4]
=(k+1)(k^2+4k+4)
=(k+1)(k+2)(k+2)
=(k+1)(k+2)^2
=(k+1)[k(k+1)+(3k+4)]
=(k+1)[k^2+k+3k+4]
=(k+1)(k^2+4k+4)
=(k+1)(k+2)(k+2)
=(k+1)(k+2)^2
2007-10-30 5:52 am
k(k+1)^2 + (k+1)(3k+4)
=k^3+2k^2+k+3k^2+7k+4
=k^3+5k^2+8k+4
=k^3+2k^2+k+3k^2+7k+4
=k^3+5k^2+8k+4
參考: 善於使用高等數學技術, 幾個步驟, 一劍封殺
收錄日期: 2021-04-23 19:36:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071029000051KK04147