Cho $0\le x,y\le 1$ thỏa mãn...

Ta có: ${{2017}^{1-x-y}}=\dfrac{{{x}^{2}}+2018}{{{y}^{2}}-2y+2019}\Leftrightarrow {{2017}^{1-y}}.\left[ {{\left( y-1 \right)}^{2}}+2018 \right]={{2017}^{x}}.\left( {{x}^{2}}+2018 \right)$
Với $0\le x,y\le 1\Rightarrow x, 1-y\in \left[ 0;1 \right]$
Xét hàm số $f(t)={{2017}^{t}}.\left( {{t}^{2}}+2018 \right)$ với $t\ge 0$ ta có:
$f'(t)=2017\ln 2017\left( {{t}^{2}}+2018 \right)+2t{{.2017}^{t}}>0$ với mọi t > 0.
Do đó $f(1-y)=f(x)\Leftrightarrow 1-y=x\Leftrightarrow x+y=1$
Ta có: $S=\left( 4{{x}^{2}}+3y \right)\left( 4{{y}^{2}}+3x \right)+25xy=16{{x}^{2}}{{y}^{2}}+12\left( {{x}^{3}}+{{y}^{3}} \right)+9xy+25xy$
$=16{{x}^{2}}{{y}^{2}}+12\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)+34xy$
$=16{{x}^{2}}{{y}^{2}}+12\left[ {{\left( x+y \right)}^{2}}-3xy \right]+34xy=16{{x}^{2}}{{y}^{2}}+12\left( 1-3xy \right)+28xy=16{{x}^{2}}{{y}^{2}}-2xy+12$
Ta có: $0\le x,y\le 1$ nên theo BĐT AM-GM thì $x+y\ge 2\sqrt{xy}\Rightarrow 0\le xy\le \dfrac{1}{4}$
Xét hàm số $f(t)=16{{t}^{2}}-2t+12\Rightarrow f'(t)=32t-2=0\Leftrightarrow t=\dfrac{1}{16}.$
Lại có: $f(0)=12;f\left( \dfrac{1}{16} \right)=\dfrac{191}{16};f\left( \dfrac{1}{4} \right)=\dfrac{25}{2}\Rightarrow \left[ \begin{aligned}
& M=\dfrac{25}{2} \\
& m=\dfrac{191}{16} \\
\end{aligned} \right.\Rightarrow M+m=\dfrac{391}{16}.$