a: \(\left\{{}\begin{matrix}x-2\sqrt{2}y=\sqrt{3}\\\sqrt{2}x+y=1-\sqrt{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2\sqrt{2}y=\sqrt{3}\\y=1-\sqrt{6}-x\sqrt{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2\sqrt{2}\left(1-\sqrt{6}-x\sqrt{2}\right)=\sqrt{3}\\x-2\sqrt{2}y=\sqrt{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2\sqrt{2}+2\sqrt{12}+4x=\sqrt{3}\\x-2\sqrt{2}y=\sqrt{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x=\sqrt{3}-4\sqrt{3}+2\sqrt{2}=-3\sqrt{3}+2\sqrt{2}\\x-2\sqrt{2}\cdot y=\sqrt{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{-3\sqrt{3}+2\sqrt{2}}{5}\\2\sqrt{2}y=x-\sqrt{3}=\dfrac{-3\sqrt{3}+2\sqrt{2}-5\sqrt{3}}{5}=\dfrac{-8\sqrt{3}+2\sqrt{2}}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{-3\sqrt{3}+2\sqrt{2}}{5}\\y=\dfrac{-8\sqrt{3}+2\sqrt{2}}{5\cdot2\sqrt{2}}=\dfrac{2\sqrt{2}\left(-2\sqrt{6}+1\right)}{5\cdot2\sqrt{2}}=\dfrac{-2\sqrt{6}+1}{5}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}x-y\sqrt{3}=0\\x\sqrt{3}+2y=1+\sqrt{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=y\sqrt{3}\\y\sqrt{3}\cdot\sqrt{3}+2y=1+\sqrt{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=y\sqrt{3}\\5y=1+\sqrt{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+1}{5}\\x=y\cdot\sqrt{3}=\dfrac{3+\sqrt{3}}{5}\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}\sqrt{2}x-\sqrt{5}y=1\\x+\sqrt{5}y=\sqrt{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\sqrt{5}=x\sqrt{2}-1\\x+x\sqrt{2}-1=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(1+\sqrt{2}\right)=\sqrt{2}+1\\y\sqrt{5}=x\sqrt{2}-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y\sqrt{5}=\sqrt{2}-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{\sqrt{2}-1}{\sqrt{5}}\end{matrix}\right.\)
d: \(\left\{{}\begin{matrix}\sqrt{2}x+\sqrt{5}y=2\\x+\sqrt{5}y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{5}\cdot y=2-x\\\sqrt{2}x+2-x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\sqrt{5}=2-x\\x\left(\sqrt{2}-1\right)=0\end{matrix}\right.\)
=>x=0 và \(y=\dfrac{2}{\sqrt{5}}\)
