$f(x)=x^2-2mx+6m-8<0$
Để pt vô nghiệm thì $f(x)≥0$
ycbt: $\Leftrightarrow \left\{ \begin{array}{l}\Delta≤0\\a>0\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l}4m^2-24m+32≤0\\1>0(ld)\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l}2≤m≤4\\1>0(ld)\end{array} \right.$
Vậy $2≤m≤4$ thoả ycbt