maths題'

2007-11-01 4:43 am
1.
6a(p+3)-10(p+3)^2

2
(4x+y)^2-(2x-y)^2

3.
(5x+y)^2-9(3x-y)^2

thx啦!

回答 (2)

2007-11-01 4:59 am
✔ 最佳答案
1.
6a(p+3) - 10(p+3)^2
= [ 6a - 10(p+3) ] (p+3)
= (6a - 10p-30) (p+3)


2
(4x+y)^2 - (2x-y)^2
= [ (4x+y) - (2x-y) ] [ (4x+y) + (2x-y) ]
= (4x + y - 2x + y) (4x + y + 2x - y)
= (2x + 2y) (6x)
= 12(x + y) x

3.
(5x+y)^2 - 9(3x-y)^2
= (5x+y)^2 - (3(3x-y))^2
= [ (5x+y) - 3(3x-y) ] [ (5x+y) + 3(3x-y) ]
= (5x + y - 9x + 3y) (5x + y + 9x - 3y)
= (4y - 4x) (14x - 2y)
= 8(y - x) (7x - y)


2007-10-31 21:01:30 補充:
上面位第三條都錯左, 記得投我一票, 做得清楚, 一步一步
2007-11-01 5:00 am
1.6a(p+3)-10(p+3)^2
=[6ap+18a-10p-30]^2
=12ap+36a-20p-60

2.(4x+y)^2-(2x-y)^2
=8x+2y-4x+2y
=4x+4y

3.(5x+y)^2-9(3x-y)^2
=10x+2y(-27x+9y)^2
=10x+2y-54x+18y
=-44+20y
參考: ME~~~如果唔啱敬請原諒=.=


收錄日期: 2021-04-23 17:19:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071031000051KK03617

檢視 Wayback Machine 備份