[(2x-1)/3]∈Z
1∈Z , deci x∈Z (0)
aplicam propietatea
[x]≤x<[x]+1
incazul nostru
[(2x-1)/3]≤(2x-1)/3<[(2x-1)/3]+1
si tinem contde egalitatea data
x+1≤(2x-1)/3<x+1+1=x+2
3x+3≤2x-1<3x+6
3x+3≤2x-1
x≤-4
2x-1<3x+6
0<x+7
-7<x
x>-7 (2)
(1)∩(2)⇒x∈(-7;-4]
dar x∈Z ( conditia (0)
deci x∈{-6;-5;-4}
verificare
pt x=-4
[-9/3]=-4+1 verifica
pt x=-5
[-11/3]=[-3,(6)]=-4 adevarat
pt x=-6
[-13/3]=[-4,(3)]=-5 adevarat
