Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
\left( {x + 2y + 1} \right)\left( {x + 2y + 2} \right) = 0\left( 1 \right)\\
xy + {y^2} + 3y + 1 = 0\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Rightarrow \left[ \begin{array}{l}
x + 2y + 1 = 0\\
x + 2y + 2 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 2y - 1\\
x = - 2y - 2
\end{array} \right.\\
+ Khi:x = - 2y - 1\\
\Rightarrow \left( { - 2y - 1} \right).y + {y^2} + 3y + 1 = 0\\
\Rightarrow - 2{y^2} - y + {y^2} + 3y + 1 = 0\\
\Rightarrow {y^2} - 2y - 1 = 0\\
\Rightarrow {y^2} - 2y + 1 = 2\\
\Rightarrow {\left( {y - 1} \right)^2} = 2\\
\Rightarrow \left[ \begin{array}{l}
y = 1 + \sqrt 2 \\
y = 1 - \sqrt 2
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = - 2y - 1 = - 3 - 2\sqrt 2 \\
x = - 3 + 2\sqrt 2
\end{array} \right.\\
+ Khi:x = - 2y - 2\\
\Rightarrow \left( { - 2y - 2} \right)y + {y^2} + 3y + 1 = 0\\
\Rightarrow - 2{y^2} - 2y + {y^2} + 3y + 1 = 0\\
\Rightarrow {y^2} - y - 1 = 0\\
\Rightarrow {y^2} - 2.\dfrac{1}{2}.y + \dfrac{1}{4} = \dfrac{5}{4}\\
\Rightarrow {\left( {y - \dfrac{1}{2}} \right)^2} = \dfrac{{\sqrt 5 }}{2}\\
\Rightarrow \left[ \begin{array}{l}
y = \dfrac{{1 + \sqrt 5 }}{2} \Rightarrow x = - 2y - 2 = - 3 - \sqrt 5 \\
y = \dfrac{{1 - \sqrt 5 }}{2} \Rightarrow x = - 3 + \sqrt 5
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( { - 3 \pm 2\sqrt 2 ;1 \mp \sqrt 2 } \right);\left( { - 3 \pm \sqrt 5 ;\dfrac{{1 \mp \sqrt 5 }}{2}} \right)} \right\}
\end{array}$
