Do \(2x^2+x+1=2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}>0;\forall x\) nên BPT tương đương:
\(\left(5-m\right)x^2-2\left(m+1\right)x+1< 0\)
Để BPT vô nghiệm
\(\Leftrightarrow\left(5-m\right)x^2-2\left(m+1\right)x+1\ge0\) đúng với mọi x
\(\Leftrightarrow\left\{{}\begin{matrix}5-m>0\\\Delta'=\left(m+1\right)^2-\left(5-m\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< 5\\m^2+3m-4\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< 5\\-4\le m\le1\end{matrix}\right.\) \(\Rightarrow-4\le m\le1\)