F3 maths一問..

2008-09-08 5:55 am
Factorization.3題數...

1. 28(x+y)^2- 7(x-y)^2

2. 4c^2- 9d^2- 2c- 3d

3. 12v- 12v^2- 6u + 3u^2

唔該大家...請盡快回答/...!!
更新1:

急ar~~!!!

更新2:

真係唔該哂2位... 2位ge答案都相當之詳細同準...!!/... 唔該哂!!... 至於最佳回答...我想問多少少問題... 點先可以學好d factorization/ factorization有咩技巧同有咩要注意??

更新3:

唉.... 3個都咁好....交投票la~~

回答 (2)

2008-09-08 6:20 am
✔ 最佳答案
1.
28(x+y)2- 7(x-y)2

= 7{4(x+y)2 - (x-y)2}

= 7{[2(x+y)]2 - (x-y)2}

= 7{2(x+y) - (x-y)}{2(x+y) + (x-y)}

= 7{2x + 2y - x + y}{2x + 2y + x - y}

= 7(x + 3y)(3x + y)

=====
2.
4c2 - 9d2 - 2c - 3d

= [4c2 - 9d2] - (2c + 3d)

= [(2c)2 - (3d)2] - (2c + 3d)

= (2c + 3d)(2c - 3d) - (2c + 3d)(1)

= (2c + 3d)[(2c - 3d) - 1]

= (2c + 3d)(2c - 3d - 1)

=====
3.
12v - 12v2 - 6u + 3u2

= 12v - 6u - 12v2 + 3u2

= (12v - 6u) - (12v2 - 3u2)

= 6(2v - u) - 3(4v2 - u2)

= 6(2v - u) - 3(2v - u)(2v + u)

= (2v - u)[6 - 3(2v + u)]

= (2v - u)(6 - 6v - 3u)

= 3(2v - u)(2 - 2v - u)
===
2008-09-08 6:22 am
1. 28(x+y)^2- 7(x-y)^2
=7[4(x+y)^2-(x-y)^2]
=7[2(x+y)+(x-y)][2(x+y)-(x-y)]
=7(3x+y)(x+3y)

2. 4c^2- 9d^2- 2c- 3d
=(2c+3d)(2c-3d)-(2c+3d)
=(2c+3d)(2c-3d-1)

3. 12v- 12v^2- 6u + 3u^2
=3(u^2-4v^2)-6(2v-u)
=3(u+2v)(u-2v)-6(2v-u)
=3(u+2v)(u-2v)+6(u-2v)
=3(u-2v)(u+2v+2)
參考: Not so hard~


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