\(\left\{{}\begin{matrix}2x+\sqrt{3}y=\sqrt{3}\\\sqrt{2}x-3y=\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x+\sqrt{3}y=\sqrt{3}\\2x-3\sqrt{2}y=2\end{matrix}\right.\)(đk: x;y\(\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}y\left(\sqrt{3}+3\sqrt{2}\right)=\sqrt{3}-\sqrt{2}\\2x+\sqrt{3}y=\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-9+4\sqrt{6}}{15}\\x=\frac{4\sqrt{3}-2\sqrt{2}}{5}\end{matrix}\right.\left(tm\right)\)
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