\(x^2-5x+6\le0\Rightarrow2\le x\le3\)
\(\left|x-m\right|>1\Rightarrow\left[{}\begin{matrix}x-m>1\\x-m< -1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m< x-1\\m>x+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m< 2\\m>3\end{matrix}\right.\)
\(\Rightarrow\) Để hệ vô nghiệm thì \(2\le m\le3\)