Gọi $S$ là số giá trị $m$ nguyên thuộc khoảng $(-20 ; 20)$ để đồ...

Vì $f(0)=48>0$ nên hàm số $y=|f(x)|$ đồng biến trên khoảng $(0 ; 2)$ khi và chỉ khi $f^{\prime}(x) \geq$ $0, \forall x \in(0 ; 2)$
$
\begin{aligned}
& \Leftrightarrow 8 x^3-12(m+4) x^2+6 m^2 x^2 \geq 0, \quad \forall x \in(0 ; 2) \\
& \Leftrightarrow g(x)=4 x^2-6(m+4) x+3 m^2 \geq 0, \quad \forall x \in(0 ; 2)
\end{aligned}
$
$\left[ \begin{aligned}
& \Delta '\le 0 \\
& \left\{ \begin{aligned}
& \Delta '>0 \\
& g\left( 0 \right)\ge 0 \\
& g\left( 2 \right)\ge 0 \\
& \left[ \begin{aligned}
& \dfrac{S}{2}\ge 2 \\
& \dfrac{S}{2}\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& -3\left( -48-24m+{{m}^{2}} \right)\le 0 \\
& \left\{ \begin{aligned}
& -3\left( -48-24m+{{m}^{2}} \right)>0 \\
& {{m}^{2}}\ge 0 \\
& 3{{m}^{2}}-12m-32\ge 0 \\
& \left[ \begin{aligned}
& \dfrac{3m+4}{4}\ge 2 \\
& \dfrac{3m+4}{4}\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left[ \begin{aligned}
& m\le 12-8\sqrt{3} \\
& m\ge 12+8\sqrt{3} \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& 12-8\sqrt{3}<m<12+8\sqrt{3} \\
& \left[ \begin{aligned}
& m\ge \dfrac{6+2\sqrt{33}}{3} \\
& m\le \dfrac{6-2\sqrt{33}}{3} \\
\end{aligned} \right. \\
& \left[ \begin{aligned}
& m\ge \dfrac{4}{3} \\
& m\le -\dfrac{4}{3} \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\begin{aligned} & {\left[\begin{array}{l}m \geq \dfrac{6+2 \sqrt{33}}{3} \\ m \leq \dfrac{6-2 \sqrt{33}}{3}\end{array}\right.} \\ & \Rightarrow m \in\{-19 ;-18 ; \ldots ;-2\} \cup\{6 ; 7 ; 8 ; \ldots ; 19\} \\ & \text { Suy ra } S=32 \text {. Vậy } S \text { chia hết cho } 4 \text {. } \\ & \end{aligned}$