Răspuns :
a,(a+1),(a+2),(a+3)=?
(a+1)+(a+2)+(a+3)=0⇒3a+6=0⇒a=-6/3⇒a=-2;
cele patru numere intregi consecutive: -2, -1, 0,1
(a+1)+(a+2)+(a+3)=0⇒3a+6=0⇒a=-6/3⇒a=-2;
cele patru numere intregi consecutive: -2, -1, 0,1
fie a, a+1,a+2,a+3 4 numere intregi consecutive;
a+1+a+2+a+3= 0, 3a+6=0, a= -2
deci numere sunt: -2,-1,0,1
a+1+a+2+a+3= 0, 3a+6=0, a= -2
deci numere sunt: -2,-1,0,1