用因式分解法解方程 (1) (x-1)눀-2(x-1)=0

2024-12-05 03:56:01
推荐回答(4个)
回答1:

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

回答2:

(x-1)²-2(x-1)=0
(x-1)[(x-1)-2]=0
(x-1)(x-3)=0
x=1 or x=3

(x+2)²-9=0
[(x+2)+3][(x+2)-3]=0
(x+5)(x-1)=0
x=-5 or x=1

回答3:

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

2(t-1)²+t=1
2(t-1)²+(t-1)=0
(t-1)[2(t-1)+1]=0
t=1或t=1/2

回答4:

(X-1)(X-1-2)=0
(X+2+3)(X+2-3)=0
[3(x-2)+2(x+1)][3(x-2)-2(x+1)=0

(t-1)[2(t-1)+1]=0