Giải HPT $\ \left\{ \begin{array}{l} 2 + 6y = \frac{x}{y} - \sqrt {x - 2y} \,\left( 1 \right)\\ \sqrt {x + \sqrt {x - 2y} } = x + 3y - 2\left( 2 \right) \end{array} \right.$

Đề bài: Giải HPT $\ \left\{ \begin{array}{l} 2 + 6y = \frac{x}{y} - \sqrt {x - 2y} \,\left( 1 \right)\\ \sqrt {x + \sqrt {x - 2y} }  = x + 3y - 2\left( 2 \right) \end{array} \right.$
Giải:
Ta có: $\ \left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l} 2y + 6{y^2} = x - y\sqrt {x - 2y} \\ y \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \left( {x - 2y} \right) - y\sqrt {x - 2y}  - 6{y^2} = 0\\ y \ne 0 \end{array} \right.$

$\  \Leftrightarrow \left\{ \begin{array}{l} {\left( {\frac{{\sqrt {x - 2y} }}{y}} \right)^2} - \frac{{\sqrt {x - 2y} }}{y} - 6 = 0\\ y \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {t^2} - t - 6 = 0\\ t = \frac{{\sqrt {x - 2y} }}{y} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} t = \frac{{\sqrt {x - 2y} }}{y} = 3\\ t = \frac{{\sqrt {x - 2y} }}{y} =  - 2 \end{array} \right.$

• Xét $\ \sqrt {x - 2y}  = 3y\left( {x + 3y} \right) - \sqrt {x + 3y}  - 2 = 0$

         $\  \Leftrightarrow \left\{ \begin{array}{l} t = \sqrt {x + 3y} \left( {t \ge 0} \right)\\ {t^2} - t - 2 = 0 \end{array} \right. \Leftrightarrow t = \sqrt {x + 3y}  = 2 \Rightarrow \left\{ \begin{array}{l} x + 3y = 4\\ \sqrt {x - 2y}  = 3y \end{array} \right.$
        $\  \Leftrightarrow \left\{ \begin{array}{l} x = 4 - 3y\\ \sqrt {4 - 5y}  = 3y \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 4 - 3y\\ 0 < y \le \frac{4}{5}\\ 4 - 5y = 9{y^2} \end{array} \right. \Leftrightarrow \left( {x;y} \right) = \left( {\frac{8}{3};\frac{4}{9}} \right).$

• Xét $\ \sqrt {x - 2y}  =  - 2y\left\{ \begin{array}{l}\sqrt {x - 2y}  = x + 3y - 2\\\sqrt {x - 2y}  =  - 2y\end{array} \right.$

         $\  \Leftrightarrow \left\{ \begin{array}{l} x = 2 - 5y\\ y < 0\\ 4{y^2} + 7y - 2 = 0 \end{array} \right. \Leftrightarrow \left( {x;y} \right) = \left( {12; - 2} \right).$
                                        Vậy $\ S = \left\{ {\left( {12; - 2} \right),\left( {\frac{8}{3};\frac{4}{9}} \right)} \right\}.$