Cho hàm số $y=\left| \dfrac{{{x}^{2}}-2mx+1}{{{x}^{2}}-x+2}...

Ta có:
$\begin{aligned}
& Max y\le 4\Leftrightarrow \left| \dfrac{{{x}^{2}}-2mx+1}{{{x}^{2}}-x+2} \right|\le 4 \forall x\in \mathbb{R} \\
& \Rightarrow \left| {{x}^{2}}-2mx+1 \right|\le 4.\left( {{x}^{2}}-x+2 \right)\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-2mx+1\le 4.\left( {{x}^{2}}-x+2 \right) \\
& -4.\left( {{x}^{2}}-x+2 \right)\le {{x}^{2}}-2mx+1 \\
\end{aligned} \right. \\
& \Leftrightarrow \left\{ \begin{aligned}
& -3{{x}^{2}}+\left( 4-2m \right)x-7\le 0 \\
& 5{{x}^{2}}-\left( 4+2m \right)x+9\ge 0 \\
\end{aligned} \right. \forall x\in R \\
& \Rightarrow \left\{ \begin{aligned}
& {{\left( 2-m \right)}^{2}}-21\le 0 \\
& {{\left( 2+m \right)}^{2}}-45\le 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -2\le m\le 6 \\
& -8\le m\le 4 \\
\end{aligned} \right.\Leftrightarrow -2\le m\le 4 \\
\end{aligned}$
Ta thấy $m\in \left[ -2;4 \right]$ thì $Max y\le 4$ nên $m\in \left( -\infty ;-2 \right]\cup \left[ 4;+\infty \right)$ thì $Max y\ge 4$.
Kết hợp với điều kiện $\left\{ \begin{aligned}
& m\in \left[ -10;10 \right] \\
& m\in \mathbb{Z} \\
\end{aligned} \right.\Rightarrow m\in \left\{ -8;-7;-6;-5;-4;-3;-2;4;5;6;7;8;9;10 \right\}$