國二數學因式分解問題 只有兩題 感恩！ 2(x-3)^2+44(x-3)+242=? 3x^2-3y^2+2x-2y=?
回答 (2)
2015-11-23 7:04 pm
✔ 最佳答案
2(x-3)^2+44(x-3)+242=2[(x-3)^2+22(x-3)+121]=
2[(x-3)+11]^2=
2(x+8)^2
3x^2-3y^2+2x-2y=
3(x^2-y^2)+2(x-y)=
3(x+y)(x-y)+2(x-y)=
[3(x+y)+2](x-y)=
(3x+3y+2)(x-y)
2015-12-04 9:00 am
2(x-3)^2+44(x-3)+242
=2[(x-3)^2+22(x-3)+121]
=2[(x-3)+11]^2
=2(x+8)^2
3x^2-3y^2+2x-2y
=3(x^2-y^2)+2(x-y)
=3(x-y)(x+y)+2(x-y)
=(x-y)(3x+3y+2)
=2[(x-3)^2+22(x-3)+121]
=2[(x-3)+11]^2
=2(x+8)^2
3x^2-3y^2+2x-2y
=3(x^2-y^2)+2(x-y)
=3(x-y)(x+y)+2(x-y)
=(x-y)(3x+3y+2)
收錄日期: 2021-04-18 14:04:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151123074404AAIiZJZ