Đáp án: $\left( {x;y} \right) = \left\{ {\left( {11;4} \right);\left( { - 1;2} \right);\left( { - 5; - 2} \right);\left( {7;4} \right)} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
{x^2} + 3{y^2} - 4xy - 2x + 6y + 5 = 0\\
\Rightarrow {x^2} + 4{y^2} + 1 + 2.x.\left( { - 1} \right) - 2.x.\left( { - 2y} \right) + 4y\\
- {y^2} + 2y + 4 = 0\\
\Rightarrow {\left( {x - 2y - 1} \right)^2} - \left( {{y^2} - 2y + 1} \right) + 5 = 0\\
\Rightarrow {\left( {x - 2y - 1} \right)^2} - {\left( {y - 1} \right)^2} = - 5\\
\Rightarrow \left( {x - 2y - 1 - y + 1} \right)\left( {x - 2y - 1 + y - 1} \right) = - 5\\
\Rightarrow \left( {x - 3y} \right)\left( {x - y - 2} \right) = - 5 = \left( { - 1} \right).5 = 1.\left( { - 5} \right)\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 3y = - 1\\
x - y - 2 = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3y = 5\\
x - y - 2 = - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3y = 1\\
x - y - 2 = - 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3y = - 5\\
x - y - 2 = 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 11;y = 4\\
x = - 1;y = 2\\
x = - 5;y = - 2\\
x = 7;y = 4
\end{array} \right.
\end{array}$
