Sửa đề: \(\left\{{}\begin{matrix}x\sqrt{3}-y=-\sqrt{3}\\3x-y\sqrt{2}=2-2\sqrt{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-\sqrt{3}y=-3\\3x-y\sqrt{2}=2-2\sqrt{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y\sqrt{3}-3x+y\sqrt{2}=-3-2+2\sqrt{6}\\x\sqrt{3}-y=-\sqrt{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-\sqrt{3}+\sqrt{2}\right)=2\sqrt{6}-5\\x\sqrt{3}=y-\sqrt{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{2\sqrt{6}-5}{-\sqrt{3}+\sqrt{2}}=\dfrac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}=\dfrac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\sqrt{2}-\sqrt{3}}=\sqrt{2}-\sqrt{3}\\x=\dfrac{y-\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{2}-2\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{6}-6}{3}\end{matrix}\right.\)