因式分解
(x^2-x)^2-4(x^2-x)-12
中二數題=====(x^2-x)^2-4(x^2-x)-12
2007-02-25 10:12 pm
回答 (5)
2007-02-25 10:17 pm
✔ 最佳答案
(x^2-x)^2-4(x^2-x)-12 =(x^2-x-6)(x^2-x+2)
=(x-3)(x+2)(x-2)(x+1)
2007-02-26 19:02:14 補充：
******注意∵-6(x^2-x)+2(x^2-x)=-4(x^2-x)∴(x^2-x)^2-4(x^2-x)-12 =(x^2-x-6)(x^2-x+2)----------------------------------------------------∵-3x+2x=-x...-2x+x=-x∴(x^2-x-6)(x^2-x+2)=(x-3)(x+2)(x-2)(x+1)
2007-02-25 10:24 pm
(x^2-x)^2-4(x^2-x)-12
=[(x^2-x)-6] [(x^2-x)+2]
=[(x-3)(x+2)] [x^2-x+2]
=[(x^2-x)-6] [(x^2-x)+2]
=[(x-3)(x+2)] [x^2-x+2]
2007-02-25 10:20 pm
(x^2-x)^2-4(x^2-x)-12
let =y
then
(x^2-x)^2-4(x^2-x)-12
=y^2-4y-12
=(y-6)(y-2)
=(x^2-x-6) 乘(x^2-x-2)
=(x-3)(x-2)乘(x-2)(x+1)
=(x-3)(x+1)(x-2)^2
2007-02-25 14:22:53 補充：
樓上correct i am wrong,should be (x-3)(x-2)(x^2-x 2)
let =y
then
(x^2-x)^2-4(x^2-x)-12
=y^2-4y-12
=(y-6)(y-2)
=(x^2-x-6) 乘(x^2-x-2)
=(x-3)(x-2)乘(x-2)(x+1)
=(x-3)(x+1)(x-2)^2
2007-02-25 14:22:53 補充：
樓上correct i am wrong,should be (x-3)(x-2)(x^2-x 2)
收錄日期: 2021-04-22 00:13:54
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070225000051KK02547