Đáp án:
$\begin{array}{l}
{x^2} - 2xy + 2{y^2} - 2x + 6y + 5 = 0\\
\Rightarrow {x^2} + {y^2} + 1 + 2.x.\left( { - 1} \right) + 2.x.\left( { - y} \right) + 2.\left( { - y} \right).\left( { - 1} \right)\\
+ {y^2} + 4y + 4 = 0\\
\Rightarrow {\left( {x - y - 1} \right)^2} + {\left( {y + 2} \right)^2} = 0\\
Do:{\left( {x - y - 1} \right)^2},{\left( {y + 2} \right)^2} \ge 0\forall x,y\\
\Rightarrow \left\{ \begin{array}{l}
x - y - 1 = 0\\
y + 2 = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = y + 1\\
y = - 2
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = - 1\\
y = - 2
\end{array} \right.
\end{array}$
Vậy x=-1 và y=-2.
