Cho hàm số $y=\left| {{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+m \right|$...

Xét $f\left( x \right)={{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+m$ trên đoạn $\left[ -1;2 \right]$.
$\Rightarrow f'\left( x \right)=4{{x}^{3}}-6{{x}^{2}}+2x;f'\left( x \right)=0\Leftrightarrow x=0;x=1;x=\dfrac{1}{2}.$
Ta có: $f\left( 0 \right)=m;f\left( \dfrac{1}{2} \right)=m+\dfrac{1}{16};f\left( -1 \right)=f\left( 2 \right)=m+4$.
Suy ra $\left\{ \begin{aligned}
& \underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=f\left( 2 \right)=m+4 \\
& \underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=f\left( 0 \right)=f\left( 1 \right)=m \\
\end{aligned} \right.$.
TH1: Nếu $m\ge 0\Rightarrow \left\{ \begin{aligned}
& m\ge 0 \\
& m+m+4=20 \\
\end{aligned} \right.\Leftrightarrow m=8.$
TH2: Nếu $m\le -4\Rightarrow \left\{ \begin{aligned}
& m\le -4 \\
& -\left( m+4 \right)-m=20 \\
\end{aligned} \right.\Leftrightarrow m=-12.$
TH3: Nếu $-4<m<0\Rightarrow \underset{\left[ -1;2 \right]}{\mathop{\min }} y=0;\underset{\left[ -1;2 \right]}{\mathop{\max }} y=\max \left\{ \left| m+4 \right|,\left| m \right| \right\}=\max \left\{ m+4,-m \right\}$.
Suy ra $\underset{\left[ -1;2 \right]}{\mathop{\min }} y+\underset{\left[ -1;2 \right]}{\mathop{\max }} y<4<0+20=20$ (loại).
Vậy tổng các giá trị của m là $-4$.