附加數.....2次方程問題 請各高手幫幫手

(3x+7) (x-1) = 4(x-1)

3x^2 + 4x - 7 = 4x -4
3x^2 - 3 = 0
x=1 or -1

2x^2-5x+3 用配方法@@

2(x^2 - 2.5x + 1.5)
=2[x^2 - 2.5x + (1.25)^2 - (1.25)^2 + 1.5]
=2(x-1.25)^2 -2(0.0625)
=2(x-1.25)^2 - 0.125

所以a= 2 , b= -1.25(-5/4) , c=0.125(1/8)

Amath來嫁咩= =?

2008-07-10 19:47:02 補充：
3x^2-11x-40 除 2x-1 = x-5

即( 2x-1 )( x-5) = 3x^2 -11x - 40

2x^2 -11x +5 = 3x^2 -11x - 40
x^2 -45 = 0
x=sqrt(45)
x=3sqrt(5)

2008-07-10 19:50:32 補充：
喂!你所講既答案,問題應該係

3x^2-11x-40 除 2x+1 = x-5

是 2x 加 1 ,不是2x 減 1
即

3x^2-11x-40 除 2x+1 = x-5
(2x+1)(x-5) = 3x^2-11x-40
2x^2 - 9x -5 = 3x^2-11x-40
x^2 -2x -35=0
(x-7)(x+5)

所以x=7 or -5

2008-07-10 19:51:30 補充：
\_/
想話就話人啦,做咩要加個令仔上去
最唔鍾意人咁講

2008-07-10 19:51:58 補充：
為左果10分仲要受你氣

2008-07-10 20:08:45 補充：
頭先我語氣也重了一些,對不起!

http://www.photo-host.org/img/449004screenhunter_01_jul._10_23.12.gif

(3x+7) (x-1) = 4(x-1)
(3x+7) (x-1) - 4(x-1) = 0
(x-1)[(3x+7) - 4] = 0
(x-1)(3x+7 - 4) = 0
(x-1)(3x+3) = 0
x = 1 or x = -1

2008-07-10 19:27:31 補充：
2x^2-5x+3
= 2(x^2 - 5x/2) + 3
= 2[x^2 - 5x/2 + (5/4)^2 - (5/4)^2] + 3
= 2(x^2 - 5/4)^2 - 2(5/4)^2 + 3
= 2(x^2 - 5/4)^2 - 1/8

3x^2-11x-40 / 2x-1 = x-5

3x^2-11x-40 =( x-5 ) ( 2x-1)

3x^2-11x-40 = 2x^2-10x -x +5

3x^2-11x-40 = 2x^2-11x +5

x^2 - 45 = 0

x^2 = 45

x = 開方45

x = 開方(9 x 5)

x = 3x 開方5