Question

用因式分解的方法解下列方程！

(1)1/9(x+1)^2-2/3(x+1)+1=0
(2)9(x-2)^2-(x+1)^2=0
(3)x^2+8x+7=0

Answer 1

1/9(x+1)^2-2/3(x+1)+1=0
(x+1)²-6(x+1)+9=0
(x+1-3)²=0
(x-2)²=0
x=2

9(x-2)²-(x+1)²=0
(3x-6)²-(x+1)²=0
(3x-6+x+1)(3x-6-x-1)=0
(4x-5)(2x-7)=0
x=1.25，或者x=3.5

x²+8x+7=0
(x+1)(x+7)=0
x=-1，或者x=-75

Answer 2

(1)1/9(x+1)^2-2/3(x+1)+1=0
解：1/9[(x+1)²-6(x+1)+9]=0
1/9（x+1-3）²=0
(x-2)²=0
x-2=0
x=2
(2)9(x-2)^2-(x+1)^2=0
[3(x-2)+(x+1)][3(x-2)-(x+1)]=0
[3x-6+x+1][3x-6-x-1]=0
[4x-5][2x-7]=0
x=5/4或x=7/2
(3)x^2+8x+7=0
(x+7)(x+1)=0
x=-7或x=-1

Answer 3

(1)1/9(x+1)^2-2/3(x+1)+1=[1/3(1+x)]^2-2/3(x+1)+1=[1/(1+x)-1)^2=0 1/(1+x)-1=0 1/(1+x)=1 1+x=1 x=0
(2)9(x-2)^2-(x+1)^2=[3(x-2)]^2- (x+1)^2=(3x-6-x-1 )(3x-6+x+1)=0 3x-6-x-1=0 x=7/3
3x-6+x+1==0 x=5/4
(3)x^2+8x+7=(x-1)(x+7)=0 x+1=0 x=-1 x+7=0x=-7