Cho $0\le x;y\le 1$ thỏa mãn...

Ta có ${{2020}^{1-x-y}}=\dfrac{{{x}^{2}}+2021}{{{y}^{2}}-2y+2022}\Leftrightarrow {{2020}^{1-y}}.\left[ {{\left( 1-y \right)}^{2}}+2021 \right]={{2020}^{x}}.\left( {{x}^{2}}+2021 \right)$ (1)
Xét hàm số $f\left( t \right)={{2020}^{t}}.\left( {{t}^{2}}+2021 \right)$ trên $\left[ 0;1 \right].$ Ta thấy hàm $f\left( t \right)$ đồng biến trên $\left[ 0;1 \right].$
Pt (1) $\Leftrightarrow f\left( 1-y \right)=f\left( x \right)\Leftrightarrow 1-y=x$ hay $x+y=1$
Ta lại có
$\begin{aligned}
& S=16{{x}^{2}}{{y}^{2}}+12{{x}^{3}}+12{{y}^{3}}+34xy \\
& =12\left( x+y \right)\left[ {{\left( x+y \right)}^{2}}-3xy \right]+16{{x}^{2}}{{y}^{2}}+34xy \\
& =16{{\left( xy \right)}^{2}}-2xy+12 \\
\end{aligned}$
Đặt $xy=u,$ điều kiện $0\le u\le \dfrac{1}{4}.$ Khi đó, $S=g\left( u \right)=16{{u}^{2}}-2u+12$ với $u\in \left[ 0;\dfrac{1}{4} \right]$
Có $g'\left( u \right)=32u-2=0\Leftrightarrow u=\dfrac{1}{16}$
$\begin{aligned}
& g\left( 0 \right)=12 \\
& g\left( \dfrac{1}{16} \right)=\dfrac{191}{16} \\
& g\left( \dfrac{1}{4} \right)=\dfrac{25}{2} \\
\end{aligned} $ $ \Rightarrow M=\underset{\left[ 0;\dfrac{1}{4} \right]}{\mathop{max}} g\left( u \right)=\dfrac{25}{2};m=\underset{\left[ 0;\dfrac{1}{4} \right]}{\mathop{\min }} g\left( u \right)=\dfrac{191}{16}$
Suy ra $M-m=\dfrac{9}{16}.$