Trường hợp 1: m=3

=>f(x)=-2(3-2)x+3=-2x+3 không thể luôn luôn dương

=>Loại

Trường hợp 2: m<>3

\(\text{Δ}=\left(2m-4\right)^2-4m\left(m-3\right)\)

\(=4m^2-16m+16-4m^2+12m=-4m+16\)

Để f(x)>0 với mọi x thì \(\left\{{}\begin{matrix}-4m+16< 0\\m-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4m< -16\\m>3\end{matrix}\right.\Leftrightarrow m>4\)

\(f\left(x\right)=\left(m-4\right)x^2+\left(m+1\right)x+2m-1\)

\(f\left(x\right)< 0,\forall x\in R\Leftrightarrow\left\{{}\begin{matrix}a< 0\\\Delta< 0\end{matrix}\right.\)

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

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

\(\Leftrightarrow m^2+2m+1-8m^2+36m-16< 0\)

\(\Leftrightarrow-7m^2+38m-15< 0\)

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

\(KL:m\in\left(5;+\infty\right)\)

\(\text{Δ}=\left(m+1\right)^2-4\cdot2=\left(m+1\right)^2-8\)

Để f(x)>0 với mọi x thì \(\left\{{}\begin{matrix}\text{Δ}< 0\\a>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}1>0\\\left(m+1\right)^2-8< 0\end{matrix}\right.\)

=>\(\left(m+1\right)^2-8< 0\)

=>\(\left(m+1\right)^2< 8\)

=>\(-2\sqrt{2}< m+1< 2\sqrt{2}\)

=>\(-2\sqrt{2}-1< m< 2\sqrt{2}-1\)