Khai triển pt (1)
\[\begin{array}{l}
\left( {y + 1} \right)\sqrt {2x - y} - {x^2} + x + xy = 0\left( * \right)\\
t = \sqrt {2x - y} \\
\to \left( * \right) \leftrightarrow \left( {y + 1} \right)t + {t^2} - {x^2} - x + y + xy = 0\\
\leftrightarrow \left( {t + x + 1} \right)\left( {t - x + y} \right) = 0\\
\leftrightarrow \left[ \begin{array}{l}
t + x + 1 = 0 \to \sqrt {2x - y} + x + 1 = 0\\
t - x + y = 0 \to \sqrt {2x - y} - x + y = 0
\end{array} \right.\\
\leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
- \left( {x + 1} \right) \ge 0\\
2x - y = {x^2} - 2x + 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x - y \ge 0\\
2x - y = {x^2} - 2xy + {y^2}
\end{array} \right.
\end{array} \right.
\end{array}\]