[tex] [ \frac{x+1}{2} ]= -1----------\ \textgreater \ -1 \leq \frac{x+1}{2} \ \textless \ -1+1;
[/tex]
Iar a doua :
-1 apartine Z
-1<=x+1 /2 | *2 <=>-2<=x+1 <=> -3 <=x------->x>= -3
Dupa
x+1 /2 < -1+1 <=>x+1 /2 <0 <=>x+1<0 =>x < -1
x∈[ -3 ; -1)
Solutia ca sa obtinem -1 : S={ -3}
