Giải hệ phương trình \(\left\{{}\begin{matrix}x-2\sqrt{2}y=\sqrt{3}\\\sqrt{2}x+y=1+\sqrt{6}\end{matrix}\right.\) , ta được nghiệm là
\(\left(x,y\right)=\left(\dfrac{2\sqrt{2}}{5}+3,\dfrac{1}{5}\right)\).\(\left(x,y\right)=\left(\dfrac{2\sqrt{2}}{5}+3,-\dfrac{1}{5}\right)\).\(\left(x,y\right)=\left(\dfrac{2\sqrt{2}}{5}+3,\dfrac{1+\sqrt{6}}{5}\right)\).\(\left(x,y\right)=\left(\dfrac{2\sqrt{2}+\sqrt{6}}{5},\dfrac{1+\sqrt{6}}{5}\right)\).Hướng dẫn giải:\(\left\{{}\begin{matrix}x-2\sqrt{2}y=\sqrt{3}\\\sqrt{2}x+y=1+\sqrt{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\sqrt{2}y+\sqrt{3}\\\sqrt{2}\left(2\sqrt{2}y+\sqrt{3}\right)+y=1+\sqrt{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\sqrt{2}y+\sqrt{3}\\4y+\sqrt{6}+y=1+\sqrt{6}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\sqrt{2}y+3\\y=\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2\sqrt{2}}{5}+3\\y=\dfrac{1}{5}\end{matrix}\right.\).