a)x²+x+1>0
x²+x+1=0
a=1;b=1;c=1
∆=b²-4ac=1²-4×1×1=1-4=-3=>∆<0=>nu exista radacini reale
b)√(x²+x+1)≥√3/2 |()²
x²+x+1≥3/2 |×2
2x²+2x+2≥3
2x²+2x+2-3≥0
2x²+2x-1≥0
a=2;b=2;c=-1
∆=b²-4ac=2²-4×2×(-1)=4+8=12
x(1,2)=(-b±√∆)/2a=(-2±√12)/2×2
x(1)=(-2+√12)/2×2=(-2+2√3)/4=[2(-1+√3)]/4
=(-1+√3)/2=(-1+1,73)/2=0,73/2=0,36=>x(1)>0
x(2)=(-2-√12)/2×2=(-2-2√3)/4=[2(-1-√3)]/4
=(-1-√3)/2=(-1-1,73)/2=-2,73/2=-1,36=>x(2)<0
2×(0,36)²+2×(-1,36)-1≥0
2×0,12-2,72-1≥0
0,24-2,72-1≥0
-3,49≥0=> F(fals)
