展開(x+1)(x+4)和(x+2)(x+3)
因式分解(x+1)(x+2)(x+3)(x+4)-48
數學題001.......因式分解
2007-10-09 3:46 pm
回答 (2)
2007-10-09 4:39 pm
✔ 最佳答案
(x-1)(x-4)=x^2-5x+4(x-2)(x-3)= x^2-5x+6
(x-1)(x-2)(x-3)(x-4)-48
=(x^2-5x+4)(x^2-5x+6)-48
=[(x^2-5x+5)-1][(x^2-5x+5)+1]-48
=(x^2-5x+5)^2-1^2-48
=(x^2-5x+5)^2-49
=(x^2-5x+5)^2-7^2
=[(x^2-5x+5)+7] [(x^2-5x+5)-7]
=(x^2-5x+12)(x^2-5x-2)
2007-10-09 4:17 pm
(x+1)(x+4)和(x+2)(x+3)=(x+1)(x+4)和(x+2)(x+3)
(x+1)(x+2)(x+3)(x+4)-48 =(x+1)(x+2)(x+3)(x+4)-48
(x+1)(x+2)(x+3)(x+4)-48 =(x+1)(x+2)(x+3)(x+4)-48
參考: ME
收錄日期: 2021-04-11 16:26:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071009000051KK00499