\(f\left(x\right)=m\left(m+2\right)x^2+2mx+2>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta'< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\left(m+2\right)>0\\m^2-2m^2-4m< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -2\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x< -4\\x>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< -2\\x>0\end{matrix}\right.\)

Để tam thức ko đổi dấu trên R

\(\Leftrightarrow\left\{{}\begin{matrix}m\left(m+2\right)>0\\\Delta'=m^2-2m\left(m+2\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\left(m+2\right)>0\\-m^2-4m< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>0\\m< -2\end{matrix}\right.\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\)

Do pt \(f\left(x\right)=0\) luôn có nghiệm với mọi m

\(\Rightarrow\) Không tồn tại m để \(f\left(x\right)>0\) \(\forall x\in R\)

Để tam thức ko đổi dấu trên R

\(\Leftrightarrow\Delta'< 0\)

\(\Leftrightarrow\left(m+1\right)^2-\left(m^2+2\right)^2< 0\)

\(\Leftrightarrow\left(m^2+m+3\right)\left(-m^2+m-1\right)< 0\)

\(\Leftrightarrow\left(m^2+m+3\right)\left(m^2-m+1\right)>0\) (luôn đúng)

Vậy với mọi m thì \(f\left(x\right)>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a< 0\\\Delta< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m-2< 0\\\left(m-3\right)^2-\left(m-2\right)\left(m-1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 0\\m^2-6m+9-m^2+3m-2< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 0\\-3m+7< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 0\\m>\frac{7}{3}\end{matrix}\right.\) (vô lý)

=> Ko tồn tại m t/m đề bài

\(f\left(x\right)=x^2-\left(m+2\right)x+8m+1>0\) với mọi x

\(\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta< 0\end{matrix}\right.\)

\(\Leftrightarrow\left(m+2\right)^2-4\left(8m+1\right)< 0\)

\(\Leftrightarrow m^2+4m+4-32m-4< 0\)

\(\Leftrightarrow m^2-28m< 0\)

\(\Leftrightarrow0< m< 28\)

\(f\left(x\right)=\left(m-4\right)x^2+\left(m+1\right)x+2m-1< 0\text{ ∀x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}a< 0\\\Delta< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m-4< 0\\\left(m+1\right)^2-4\left(m-4\right)\left(2m-1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\m^2+2m+1-4\left(2m^2-9m+4\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\m^2+2m+1-8m^2+36m-16< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\-7m^2+38m-15< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\\left[{}\begin{matrix}m< \frac{3}{7}\\m>5\end{matrix}\right.\end{matrix}\right.\Leftrightarrow m< \frac{3}{7}\)

\(f\left(x\right)=\left(2-m\right)x^2-2x+1>0\) ∀x

\(\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2-m>0\\1-2+m< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 2\\m< 1\end{matrix}\right.\Leftrightarrow m< 1\)

\(f\left(x\right)=\left(m+2\right)x^2+2\left(m+2\right)x+m+3>0\) ∀x

\(\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta'< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m+2>0\\\left(m+2\right)^2-\left(m+2\right)\left(m+3\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>-2\\m^2+4m+4-m^2-5m-6< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>-2\\-m-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-2\\m>-2\end{matrix}\right.\)

Vậy m>-2

Để dấu tam thức ko đổi trên R

\(\Leftrightarrow\Delta'< 0\)

\(\Leftrightarrow4\left(m+1\right)^2+1-m^2< 0\)

\(\Leftrightarrow3m^2+8m+5< 0\Rightarrow-\frac{5}{3}< m< -1\)