Giải

Ta có : $5{{x}^{2}}+5{{y}^{2}}+8xy-2x+2y+2=0$

$\Leftrightarrow (4{{x}^{2}}+4{{y}^{2}}+8xy)+({{x}^{2}}-2x+1)+({{y}^{2}}+2y+1)=0$

$\Leftrightarrow 4{{(x+y)}^{2}}+{{(x-1)}^{2}}+{{(y+1)}^{2}}=0$

Vì ${{(x+y)}^{2}}\ge 0;\,\,{{(x-1)}^{2}}\ge 0;\,\,{{(y+1)}^{2}}\ge 0$ với mọi $x,y,z$ nên

$4{{(x+y)}^{2}}+{{(x-1)}^{2}}+{{(y+1)}^{2}}=0$

$\Leftrightarrow \left\{ \begin{align}& x+y=0 \\ & x-1=0 \\ & y+1=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x=1 \\ & y=-1 \\ \end{align} \right.$

$M={{(x+y)}^{2019}}+{{(x-2)}^{2020}}+{{(y+1)}^{2021}}=0+{{(1-2)}^{2020}}+0=1$