Răspuns:
caz 1
|x|>0
|x-2|>0
x+x-2=0
2x-2=0
2x=2
x=1
|x|+|x-2|>0
x+x-2>0
2x-2>0
2x>2
x>1
x={2, 3, 4, ...}
caz 2
|x|<0
|x-2|<0
-x-x+2=0
-2x+2=0
-2x=-2
x=1
|x|+|x-2|>0
-x-x+2>0
-2x+2>0
-2x>-2
x>1
x={2, 3, 4, ...}
caz 1 = caz 2
caz 3
|x|>0
|x-2|<0
x-x+2=0
2=0
imposibil
|x|+|x-2|>0
x-x+2>0
2>0
imposibil
caz 4
|x|<0
|x-2|>0
-x+x-2=0
-2=0
imposibil
|x|+|x-2|>0
-x+x-2>0
-2>0
imposibil