Ta có f(x) = x2 + x + 1 = \(\left(x^2+\frac{1}{2}x\right)+\left(\frac{1}{2}x+\frac{1}{4}\right)+\frac{3}{4}=x\left(x+\frac{1}{2}\right)+\frac{1}{2}\left(x+\frac{1}{2}\right)+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(\text{vì }\left(x+\frac{1}{2}\right)^2\ge0\forall x\right)\)
=> f(x) vô nghiệm