Conform definitiei modulului, distingem 3 cazuri
1. x>0
atunci |x-x| <2
|0|<2
0<2, adevarat oricare x>0 deci x∈(0,∞)
2. x=0
0<2 adevarat, 0 verifica
3. x<0, deci|x|= -x
|x-(-x)|<2
|-2x|<2
2 ca numar pozitiv iese
2|-x|<2
|-x|<1
|x|<1
x∈( -1,1) dar x<0
deci x∈(-1,1)∩(-∞; 0)=(-1;0)
deci in total x∈(-1;0)∪{0}∪(0,∞)=(-1,∞) cerinta
