$\begin{array}{l}
y = \frac{{x + 2}}{{x + 5m}}\\
TXD:D = R\backslash \left\{ { - 5m} \right\}\\
y' = \frac{{5m - 2}}{{{{\left( {x + 5m} \right)}^2}}}\\
HS\,dong\,bien\,tren\,\left( { - \infty ; - 10} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
5m - 2 > 0\\
- 5m \ge - 10
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m > \frac{2}{5}\\
m \le 2
\end{array} \right. \Rightarrow m \in \left\{ {1;2} \right\}
\end{array}$
Chọn A.