Giải thích các bước giải:
ĐK: \[\left\{ \begin{array}{l}
y \ge 0\\
x \ge 1
\end{array} \right.\]
Ta có:
\[\begin{array}{l}
xy + x + y = {x^2} - 2{y^2}\\
\Leftrightarrow \left( {{x^2} - xy - 2{y^2}} \right) - \left( {x + y} \right) = 0\\
\Leftrightarrow \left( {x - 2y} \right)\left( {x + y} \right) - \left( {x + y} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + y = 0\\
x - 2y = 1
\end{array} \right.
\end{array}\]
Nếu x=-y thì ta có:
\[\begin{array}{l}
\left( { - y} \right)\sqrt {2y} - y\sqrt { - y - 1} = 2.\left( { - y} \right) - 2y\\
\Rightarrow y < - 1\left( L \right)
\end{array}\]
Nếu x-2y=1 thì 2y=x-1, ta có:
\[\begin{array}{l}
x\sqrt {x - 1} - \frac{{x - 1}}{2}\sqrt {x - 1} = 2x - \left( {x - 1} \right)\\
\Leftrightarrow \sqrt {x - 1} \left( {x - \frac{{x - 1}}{2}} \right) = x + 1\\
\Leftrightarrow \sqrt {x - 1} .\frac{{x + 1}}{2} = x + 1\\
\Rightarrow \left[ \begin{array}{l}
x + 1 = 0 \Rightarrow x = - 1(L)\\
\sqrt {x - 1} = 2 \Rightarrow x = 5 \Rightarrow y = 2
\end{array} \right.
\end{array}\]
