\(f\left(x\right)=x^2+2x+3=x^2+2x+1+2=\left(x+1\right)^2+2\)
Do \(\left(x+1\right)^2\ge0\Rightarrow f\left(x\right)=\left(x+1\right)^2+2\ge2>0\)
\(\Rightarrow f\left(x\right)\) vô nghiệm
Vậy đa thức f(x) không có nghiệm
Ta có: \(f\left(x\right)=x^2+2x+3\)
\(=x^2+x+x+3\)
\(=\left(x^2+x\right)+\left(x+1\right)+2\)
\(=x\left(x+1\right)+\left(x+1\right)+2\)
\(=\left(x+1\right)\left(x+1\right)+2\)
= \(\left(x+1\right)^2+2\)
Vì \(\left(x+1\right)^2\ge0\) \(\forall x\)
\(\Rightarrow\left(x+1\right)^2+2\ge2>0\)
\(\Rightarrow f\left(x\right)>0\)
\(\Rightarrow f\left(x\right)\) vô nghiệm.
