Giải hệ phương trình:  \(\left\{ \begin{array}{l}x\left( {x - 3y} \right) + y\left( {y + x} \right) = 0\\\sqrt x .\sqrt {y - 2}  = 1\end{array} \right.\) 

Điều kiện:

\(\left\{ \begin{array}{l}x \ge 0\\y - 2 \ge 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 2\end{array} \right.\)

\(\left\{ \begin{array}{l}x\left( {x - 3y} \right) + y\left( {y + x} \right) = 0\\\sqrt x .\sqrt {y - 2}  = 1\end{array} \right.\)  \( \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 2xy + {y^2} = 0\\x\left( {y - 2} \right) = 1\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{\left( {x - y} \right)^2} = 0\\x\left( {y - 2} \right) = 1\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = y\\x\left( {y - 2} \right) = 1\,\,\,\left( * \right)\end{array} \right.\)

Thay \(x = y\) vào \(\left( * \right)\), ta có :

\(\left\{ \begin{array}{l}x = y\\y\left( {y - 2} \right) = 1\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = y\\{\left( {y - 1} \right)^2} - 2 = 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}x = y\\{\left( {y - 1} \right)^2} = 2\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = y\\\left[ \begin{array}{l}y - 1 = \sqrt 2 \\y - 1 =  - \sqrt 2 \end{array} \right.\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}x = y\\\left[ \begin{array}{l}y = \sqrt 2  + 1\\y =  - \sqrt 2  + 1\end{array} \right.\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = \sqrt 2  + 1\\y = \sqrt 2  + 1\end{array} \right.\\\left\{ \begin{array}{l}x =  - \sqrt 2  + 1\\y =  - \sqrt 2  + 1\end{array} \right.\end{array} \right.\)

Vậy hệ phương trình có hai nghiệm là:

\(\left( {x;\,\,y} \right) \in \left\{ {\left( {\sqrt 2  + 1;\,\,\sqrt 2  + 1} \right),\left( { - \sqrt 2  + 1;\,\, - \sqrt 2  + 1} \right)} \right\}.\)