Giải hệ phương trình: \(\left\{\begin{matrix} 2+6y=\frac{x}{y}-\sqrt{x-2y}

Từ phương trình (1) \((2y+\sqrt{x-2y})(3y-\sqrt{x-2y})=0\Leftrightarrow \bigg \lbrack \begin{matrix} -2y=\sqrt{x-2y}\\ 3y=\sqrt{x-2y} \end{matrix}\)
Từ (2) \(\Leftrightarrow 3^{\sqrt{x+\sqrt{x-2y}}}.2^{2x+6y-4}=2^{2.\sqrt{x+\sqrt{x-2y}}}.3^{x+3y-2}+4^{\sqrt{x+\sqrt{x-2y}}}\)
\(\Leftrightarrow 4^{x+3y-2}(3^{\sqrt{x\sqrt{x-2y}}}+1)=4^{\sqrt{x\sqrt{x-2y}}}(3^{x+3y-1}+1)\)
\(\Leftrightarrow \frac{3^{\sqrt{x\sqrt{x-2y}}}+1}{4^{\sqrt{x+\sqrt{x-2y}}}}=\frac{3^{x+3y-2}+1}{4^{x+3y-2}}\)
Ta có hàm số  \(f(t)=\frac{3^t+1}{4^t}\) là hàm nghịch biến 
\((2)\Leftrightarrow \sqrt{x+\sqrt{x-2y}}=x+3y-2\)
TH1 \(\left\{\begin{matrix} \sqrt{x+\sqrt{x-2y}}=x+3y-2\\ -2y=\sqrt{x-2y} \end{matrix}\right.\Leftrightarrow\left\{\begin{matrix} x+5y=2\\ 4y^2+2y=x\\ y\leq 0\\ x\geq 2y \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=12\\ y=-2 \end{matrix}\right.\)
\(\left\{\begin{matrix} 3y=\sqrt{x-2y}\\ \sqrt{x+\sqrt{x-2y}}=x+3y-2 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3y=\sqrt{x-2y}\\ \sqrt{x-3y}=x+3y-2 \end{matrix}\right.\)
TH2 \(\Leftrightarrow \left\{\begin{matrix} 3y=\sqrt{x-2y}\\ \sqrt{x+3y}=2 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 9y^2+2y=x\\ x+3y=4\\ y\geq 0\\ x\geq 2y \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{8}{3}\\ \\ y=\frac{4}{9} \end{matrix}\right.\)
Vậy hệ có nghiêm là (12;-2) và(\(\frac{8}{3};\frac{4}{9}\))