F.2 Maths唔識 好趕 20點 入左先

係簡易多項式的因式分解1. (2+3c)^2-(2c+4)^2A. (2+3c)^2-(2c+4)^2= (2+3c+2c+4)(2+3c-(2c+4))= (5c+6)(2+3c-2c-4)= (5c+6)(c-2)2. 81-u^4A. 81-u^4= 9^2-(u^2)^2= (3^2)^2-(u^2)^2= (3^2-u^2)(3^2+u^2)= (3-u)(3+u)(9+u^2)3. (a-b)^2+6(a-b)+9A. (a-b)^2+6(a-b)+9{= (a-b)^2+2*1*3(a-b)+3^2}--------------This is not necessary= (a-b+3)^24. 4-20(c-3d)+25(c-3d)^2A. 4-20(c-3d)+25(c-3d)^2= 25(c-3d)^2-20(c-3d)+4= 5^2(c-3d)^2-2*5*2(c-3d)+2^2= [5(c-3d)-2]^2= (5c-15d-2)^2啱先有最佳錯一條就冇

2010-12-03 20:01:21 補充：
係簡易多項式的因式分解
1. (2+3c)^2-(2c+4)^2

A. (2+3c)^2-(2c+4)^2
= (2+3c+2c+4)(2+3c-(2c+4))
= (5c+6)(2+3c-2c-4)
= (5c+6)(c-2)

2. 81-u^4

A. 81-u^4
= 9^2-(u^2)^2
= (3^2)^2-(u^2)^2
= (3^2-u^2)(3^2+u^2)
= (3-u)(3+u)(9+u^2)

2010-12-03 20:01:26 補充：
3. (a-b)^2+6(a-b)+9

A. (a-b)^2+6(a-b)+9
{= (a-b)^2+2*1*3(a-b)+3^2}--------------This is not necessary
= (a-b+3)^2

4. 4-20(c-3d)+25(c-3d)^2

A. 4-20(c-3d)+25(c-3d)^2
= 25(c-3d)^2-20(c-3d)+4
= 5^2(c-3d)^2-2*5*2(c-3d)+2^2
= [5(c-3d)-2]^2
= (5c-15d-2)^2

啱先有最佳
錯一條就冇

2010-12-08 19:00:30 補充：
Please state the questions I have done wrong.

你第四題錯左
第二題都係，但係可能打錯sor

1.
(2+3c)^2-(2c+4)^2
= [(2+3c)+(2c+4)] [(2+3c)-(2c+4)]
= (2+3c+2c+4) (2+3c-2c-4)
= (5c+6) (c-2)

2. 81-u^4
= (3^2)^2 - (u^2)^2
= (3^2+u^2) (3^2-u^2)
= (9+u^2) (3+u) (3-u)

3.
(a-b)^2+6(a-b)+9
= (a-b)^2+2(a-b)(3)+3^2
= [(a-b)+3]^2
=(a-b+3)^2

4.
4-20(c-3d)+25(c-3d)^2
= 2^2 - 2(c-3d)(5) + [5(c-3d)]^2
= [2-5(c-3d)]^2
= (2-5c+15d)^2



2010-12-03 22:23:38 補充：
4. 更正一下:
4-20(c-3d)+25(c-3d)^2
= 2^2 + 2(c-3d)(-5) + [-5(c-3d)]^2
= [2+5(c-3d)]^2
= (2+5c-15d)^2

2010-12-03 22:24:32 補充：
4.
4-20(c-3d)+25(c-3d)^2
= 2^2 - 2(2)(c-3d)(5) + [5(c-3d)]^2 <== 中間漏了*2
= [2-5(c-3d)]^2
= (2-5c+15d)^2

2010-12-03 22:26:10 補充：
4. 最後更正:
4-20(c-3d)+25(c-3d)^2
= (-2)^2 + 2(-2)[5(c-3d)] + [5(c-3d)]^2
= [-2+5(c-3d)]^2
= (5c-15d-2)^2