\(\left\{{}\begin{matrix}\sqrt{3}x-\sqrt{2}y=1\\\sqrt{2}x+\sqrt{3}y=\sqrt{3}\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\sqrt{3}x-\sqrt{2}y=1\left(1\right)\\x=\frac{\sqrt{3}-\sqrt{3}y}{\sqrt{2}}\end{matrix}\right.\)
Thay x vào (1) có:
\(\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{3}y\right)}{\sqrt{2}}-\sqrt{2}y=1\) <=> \(\frac{3-3y-2y}{\sqrt{2}}=1\) <=> \(3-5y=\sqrt{2}\)
<=> \(5y=3-\sqrt{2}\) <=> \(y=\frac{3-\sqrt{2}}{5}\)
=> x=\(\frac{\sqrt{3}-\sqrt{3}.\frac{3-\sqrt{2}}{5}}{\sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}\left(1-\frac{3-\sqrt{2}}{5}\right)=\frac{\sqrt{3}}{\sqrt{2}}.\frac{2+\sqrt{2}}{5}=\frac{\sqrt{3}}{\sqrt{2}}.\frac{\sqrt{2}\left(1+\sqrt{2}\right)}{5}=\frac{\sqrt{3}\left(1+\sqrt{2}\right)}{5}\)
