Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 2y + 1 = 2\\
{y^2} + 2x + 1 = 2
\end{array} \right.\\
\Rightarrow {x^2} - 2y + 1 - {y^2} - 2x - 1 = 2 - 2\\
\Rightarrow {x^2} - {y^2} - 2x - 2y = 0\\
\Rightarrow \left( {x - y} \right)\left( {x + y} \right) - 2\left( {x + y} \right) = 0\\
\Rightarrow \left( {x + y} \right)\left( {x - y - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = - y\\
x = y + 2
\end{array} \right.\\
+ Khi:x = - y\\
\Rightarrow {x^2} + 2x + 1 = 2\\
\Rightarrow {\left( {x + 1} \right)^2} = 2\\
\Rightarrow \left[ \begin{array}{l}
x = \sqrt 2 - 1 \Rightarrow y = 1 - \sqrt 2 \\
x = - \sqrt 2 - 1 \Rightarrow y = \sqrt 2 + 1
\end{array} \right.\\
+ khi:x = y + 2\\
\Rightarrow {y^2} + 2\left( {y + 2} \right) + 1 = 2\\
\Rightarrow {y^2} + 2y + 3 = 0\\
\Rightarrow {\left( {y + 1} \right)^2} = - 2\left( {ktm} \right)\\
Vay\,\left( {x;y} \right) = \left( {\sqrt 2 - 1;1 - \sqrt 2 } \right)/\left( { - \sqrt 2 - 1;\sqrt 2 + 1} \right)
\end{array}$
