Cho hàm sô $f\left( x \right)=\left| \dfrac{{{x}^{2}}-mx+2m}{x-2}...

Xét hàm số $g\left( x \right)=\dfrac{{{x}^{2}}-mx+2m}{x-2}\Rightarrow g'\left( x \right)=\dfrac{{{x}^{2}}-4x}{{{\left( x-2 \right)}^{2}}}=0\Rightarrow \left[ \begin{aligned}
& x=0 \\
& x=4 \\
\end{aligned} \right.$
Khi $x=0\Rightarrow g\left( 0 \right)=-m$. Ta có $g\left( -1 \right)=\dfrac{1}{3}\left( -3m-1 \right)=-m-\dfrac{1}{3};g\left( 1 \right)=\dfrac{1+m}{-1}=-1-m$
Mà $-1-m<-\dfrac{1}{3}-m<-m$
Suy ra $\underset{\!\![\!\!-1;1]}{\mathop{\max }} f\left( x \right)=\max \left\{ \left| m \right|,\left| m+1 \right|,\left| m+\dfrac{1}{3} \right| \right\}=\max \left\{ \left| m \right|,\left| m+1 \right| \right\}$
TH1: $\left\{ \begin{aligned}
& \left| m+1 \right|\ge \left| m \right| \\
& \left| m+1 \right|\le 5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m\ge -\dfrac{1}{2} \\
& -6\le m\le 4 \\
\end{aligned} \right.\Rightarrow m\in \left\{ 0;1;2;3;4 \right\}$
TH2: $\left\{ \begin{aligned}
& \left| m+1 \right|<\left| m \right| \\
& \left| m \right|\le 5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m<-\dfrac{1}{2} \\
& -5\le m\le 5 \\
\end{aligned} \right.\Rightarrow m\in \left\{ -5;-4;-3;-2;-1 \right\}$
Suy ra tổng các phần tử của S bằng – 5.