Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} = y + 2\\
{y^2} = x + 2
\end{array} \right.\\
\Rightarrow {x^2} - {y^2} = y - x\\
\Rightarrow \left( {x - y} \right)\left( {x + y} \right) + \left( {x - y} \right) = 0\\
\Rightarrow \left( {x - y} \right)\left( {x + y + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = y\left( 1 \right)\\
x + y + 1 = 0\left( 2 \right)
\end{array} \right.\\
\left( 1 \right):{x^2} = x + 2\\
\Rightarrow {x^2} - x - 2 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2 = y\\
x = - 1 = y
\end{array} \right.\\
\left( 2 \right):{x^2} = - 1 - x + 2\\
\Rightarrow {x^2} + x - 1 = 0\\
\Rightarrow x = \frac{{ - 1 \pm \sqrt 5 }}{2} \Rightarrow y = \frac{{ - 1 \pm \sqrt 5 }}{2}
\end{array}$
