Đáp án:
$\begin{array}{l}
a){x^2} + 4{y^2} - 2x + 4y + 2 = 0\\
\Leftrightarrow {x^2} - 2x + 1 + 4{y^2} + 4y + 1 = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} + {\left( {2y + 1} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {x - 1} \right)^2} \ge 0\\
{\left( {2y + 1} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 1 = 0\\
2y + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = - \dfrac{1}{2}
\end{array} \right.\\
Vậy\,x = 1;y = - \dfrac{1}{2}\\
b)2{x^2} + {y^2} - 2xy - 2y + 2 = 0\\
\Leftrightarrow {x^2} + {y^2} + 1 - 2xy - 2y + 2x\\
+ {x^2} - 2x + 1 = 0\\
\Leftrightarrow {x^2} + {\left( { - y} \right)^2} + {1^2} + 2.x.\left( { - y} \right) + 2.\left( { - y} \right).1 + 2.x.1\\
+ {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow {\left( {x - y + 1} \right)^2} + {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x - y + 1 = 0\\
x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = x + 1\\
x = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 2\\
x = 1
\end{array} \right.\\
Vậy\,x = 1;y = 2
\end{array}$
