Điều kiện $x\ge 0$
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + x{y^2} - xy - {y^3} = 0\left( 1 \right)\\
\left( {2x + 1} \right)\sqrt x + y - 1 = 0\left( 2 \right)
\end{array} \right.\\
\left( 1 \right){x^2} + x{y^2} - xy - {y^3} = 0\\
\Leftrightarrow x\left( {x - y} \right) + {y^2}\left( {x - y} \right) = 0\\
\Leftrightarrow \left( {x - y} \right)\left( {x + {y^2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = y\left( 3 \right)\\
x = - {y^2}\left( 4 \right)
\end{array} \right.\\
\left( 3 \right) \to \left( 2 \right):\left( {2x + 1} \right)\sqrt x + x - 1 = 0\\
\Leftrightarrow 2x\sqrt x + \sqrt x + x - 1 = 0\\
\Leftrightarrow 2x\sqrt x - x + 2x - \sqrt x + 2\sqrt x - 1 = 0\\
\Leftrightarrow x\left( {2\sqrt x - 1} \right) + \sqrt x \left( {2\sqrt x - 1} \right) + 2\sqrt x - 1 = 0\\
\Leftrightarrow \left( {2\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right) = 0\\
\Leftrightarrow \sqrt x = \dfrac{1}{2} \Rightarrow x = \dfrac{1}{4}\\
\to x = y = \dfrac{1}{4}\\
\left( 4 \right) \to \left( 2 \right):x = - {y^2} \to x \le 0 \Rightarrow x = 0(TM)\\
\Rightarrow y = 0\\
\to x = y = 0\\
\Rightarrow \left( {x;y} \right) = \left( {0;0} \right),\left( {\dfrac{1}{4};\dfrac{1}{4}} \right)
\end{array}$
