Question

Chứng minh rằng: x^2 + x + 1 >0, với mọi x

Answer 1

x²+x+1=x²+2.x.1/2+1/4+3/4

=(x+1/2)²+3/4

Vì (x+1/2)²\(\ge\)0 nên

(x+1/2)²+3/4>0

=>x²+x+1>0

Answer 2

x2+x+1=x2+2.x.1/2+1/4+3/4 =(x+1/2)2+3/4 Vì (x+1/2)2 ≥ 0 nên (x+1/2)2+3/4>0 =>x2+x+1>0