Đáp án:
$\begin{array}{l}
3{x^2} + {y^2} + 4xy + 4x + 2y + 5 = 0\\
\Rightarrow 4{x^2} + {y^2} + 1 + 4xy + 4x + 2y - {x^2} + 4 = 0\\
\Rightarrow {\left( {2x} \right)^2} + {y^2} + 1 + 2.2.y + 2.2x + 2.y.1 - {x^2} + 4 = 0\\
\Rightarrow {\left( {2x + y + 1} \right)^2} - {x^2} = - 4\\
\Rightarrow \left( {2x + y + 1 - x} \right)\left( {2x + y + 1 + x} \right) = - 4\\
\Rightarrow \left( {x + y + 1} \right)\left( {3x + y + 1} \right) = - 4 = \left( { - 1} \right).4 = 1.\left( { - 4} \right) = \left( { - 4} \right).1 = 4.\left( { - 1} \right) = 2.\left( { - 2} \right) = \left( { - 2} \right).2\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{5}{2};y = \frac{9}{2}\\
x = \frac{{ - 5}}{2};y = \frac{5}{2}\\
x = \frac{5}{2};y = \frac{{ - 15}}{2}\\
x = \frac{{ - 5}}{2};y = \frac{{11}}{2}\\
x = - 2;y = 3\\
x = 2;y = - 5
\end{array} \right.
\end{array}$
