Đáp án:

\(\left[ \begin{array}{l}
 - \dfrac{5}{2} \le m \le \dfrac{3}{2}\\
 - 4 < m < 4
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
\left| {\dfrac{{{x^2} + x + 4}}{{{x^2} - mx + 4}}} \right| \le 2\forall x\\
 \to \left[ \begin{array}{l}
\dfrac{{{x^2} + x + 4}}{{{x^2} - mx + 4}} \le 2\\
\dfrac{{{x^2} + x + 4}}{{{x^2} - mx + 4}} \ge  - 2
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\dfrac{{{x^2} + x + 4 - 2{x^2} + 2mx - 8}}{{{x^2} - mx + 4}} \le 0\\
\dfrac{{{x^2} + x + 4 + 2{x^2} - 2mx + 8}}{{{x^2} - mx + 4}} \ge 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\dfrac{{ - {x^2} + \left( {2m + 1} \right)x - 4}}{{{x^2} - mx + 4}} \le 0\\
\dfrac{{3{x^2} + \left( {1 - 2m} \right)x + 12}}{{{x^2} - mx + 4}} \ge 0
\end{array} \right.\\
TH1:\dfrac{{ - {x^2} + \left( {2m + 1} \right)x - 4}}{{{x^2} - mx + 4}} \le 0\\
 \to \left\{ \begin{array}{l}
 - {x^2} + \left( {2m + 1} \right)x - 4 \le 0\forall x\\
{x^2} - mx + 4 > 0\forall x
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
4{m^2} + 4m + 1 - 4.\left( { - 1} \right).\left( { - 4} \right) \le 0\\
{m^2} - 4.4 < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
4{m^2} + 4m - 15 \le 0\\
{m^2} < 16
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
 - \dfrac{5}{2} \le m \le \dfrac{3}{2}\\
 - 4 < m < 4
\end{array} \right.\\
 \to  - \dfrac{5}{2} \le m \le \dfrac{3}{2}\\
TH2:\dfrac{{3{x^2} + \left( {1 - 2m} \right)x + 12}}{{{x^2} - mx + 4}} \ge 0\forall x\\
 \to \left\{ \begin{array}{l}
3{x^2} + \left( {1 - 2m} \right)x + 12 \ge 0\forall x\\
{x^2} - mx + 4 > 0\forall x
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
1 - 4m + 4{m^2} - 4.3.12 \le 0\\
{m^2} - 4.4 < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
4{m^2} - 4m - 143 \le 0\\
 - 4 < m < 4
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
 - \dfrac{{11}}{2} \le m \le \dfrac{{13}}{2}\\
 - 4 < m < 4
\end{array} \right.\\
 \to  - 4 < m < 4\\
KL:\left[ \begin{array}{l}
 - \dfrac{5}{2} \le m \le \dfrac{3}{2}\\
 - 4 < m < 4
\end{array} \right.
\end{array}\)