Cho các số x,y thoả mãn đẳng thức:         5{x^2} + 5{y^2} + 8xy - 2x + 2y + 2 = 0.

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

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

\({\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} + 4{\left( {x + y} \right)^2} = 0\)

Vì \(\left\{ \begin{array}{l}{\left( {x - 1} \right)^2} \ge 0,\forall x\\{\left( {y + 1} \right)^2} \ge 0,\forall y\\{\left( {x + y} \right)^2} \ge 0,\forall x,y\end{array} \right.\) nên \(\left\{ \begin{array}{l}{\left( {x - 1} \right)^2} = 0\\{\left( {y + 1} \right)^2} = 0\\{\left( {x + y} \right)^2} = 0\end{array} \right.\) suy ra \(\left\{ \begin{array}{l}x - 1 = 0\\y + 1 = 0\\x + y = 0\end{array} \right.\) nên \(\left\{ \begin{array}{l}x = 1\\y = - 1\\x + y = 0\end{array} \right.\).

Ta có: \(M = {\left( {x + y} \right)^{2021}} + {\left( {x - 2} \right)^{2022}} + {\left( {y + 1} \right)^{2023}}\)

\( = {0^{2021}} + {\left( {1 - 2} \right)^{2022}} + {\left( { - 1 + 1} \right)^{2023}}\)

\( = 0 + {\left( { - 1} \right)^{2022}} + {0^{2023}}\)\( = 0 + 1 + 0 = 1.\)

Vậy \(M = 1\).