Đáp án:
$\begin{array}{l}
{x^2} + 2xy + 7\left( {x + y} \right) + {y^2} + 10 = 0\\
\Rightarrow {x^2} + 2xy + {y^2} + 7\left( {x + y} \right) + 10 = 0\\
\Rightarrow {\left( {x + y} \right)^2} + 7\left( {x + y} \right) + 10 = 0\\
Đặt:x + y = t\\
\Rightarrow {t^2} + 7t + 10 = 0\\
\Rightarrow {t^2} + 2t + 5t + 10 = 0\\
\Rightarrow \left( {t + 2} \right)\left( {t + 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
t = - 2\\
t = - 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x + y = - 2\\
x + y = - 5
\end{array} \right.\\
Do:x;y \in Z\\
\Rightarrow \left[ \begin{array}{l}
x = - 2 - y\left( {y \in Z} \right)\\
x = - 5 - y\left( {y \in Z} \right)
\end{array} \right.
\end{array}$
