b) \(f\left(x\right)=x^2+x+1\)
\(\Rightarrow f\left(x\right)=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(\Rightarrow f\left(x\right)=\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}\)
\(\Rightarrow f\left(x\right)=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\) \(\forall x\in R.\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0.\)
\(\Rightarrow f\left(x\right)>0.\)
Hay \(f\left(x\right)\ne0\)
Vậy đa thức \(f\left(x\right)\) không có nghiệm.
Chúc bạn học tốt!