Question

解不等式：|x-1|+|x-2|<=2

Answer 1

当x>2时|x-1|+|x-2|=2x-3≤2解得2<x≤2.5
当x<1时|x-1|+|x-2|=3-2x≤2解得0.5≤x<1
当1≤x≤2时|x-1|+|x-2|=1≤2∴1≤x≤2
综上所述此不等式的解集为{x{0.5≤x≤2.5}

Answer 2

讨论，当X>2,时，X-1+X-2<=2
1=<X=<2 X-1+2-X<=2
X<1 1-X+2-X<=2
三种条件，分别解，最后取并集。

Answer 3

case 1 : x<1
|x-1|+|x-2|<=2

-(x-1)-(x-2) <=2
x >=-1/2
solution for case 1: -1/2 <=x < 1

case 2 : x=1
|x-1|+|x-2| = 1 <=2 true

case 3: 1<x<2
|x-1|+|x-2|<=2

x-1-(x-2)<=2
1<=2
true
solution for case 2: 1<x<2

case 4: x=2
|x-1|+|x-2| =1 <=2 true

case 5: x>2
|x-1|+|x-2| <=2

x-1+ x-2 <=2
x<=5/2
solution for case 5 : 2 <x<=5/2

solution
case 1 or case 2 or case 3 or case 4 or case 5
-1/2<=x<=5/2

Answer 4