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 ${\left( {x;y} \right) = \left( {\frac{{ – 6 + \sqrt 2 }}{8};\frac{{ – \sqrt 2 – 1}}{4}} \right)}$ Bình luận
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 ${\left( {x;y} \right) = \left( {\frac{{ – 6 + \sqrt 2 }}{8};\frac{{ – \sqrt 2 – 1}}{4}} \right)}$