Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
x\sqrt 2 - 3y = 1\\
2x + y\sqrt 2 = - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x - 3\sqrt 2 y = \sqrt 2 \\
2x + y\sqrt 2 = - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 3\sqrt 2 y - y\sqrt 2 = \sqrt 2 - \left( { - 2} \right)\\
2x + y\sqrt 2 = - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 4\sqrt 2 y = 2 + \sqrt 2 \\
x = \frac{{ - 2 - y\sqrt 2 }}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = \frac{{ - \sqrt 2 - 1}}{4}\\
x = \frac{{ - 6 + \sqrt 2 }}{8}
\end{array} \right.
\end{array}$
Vậy ${\left( {x;y} \right) = \left( {\frac{{ - 6 + \sqrt 2 }}{8};\frac{{ - \sqrt 2 - 1}}{4}} \right)}$
