Đáp án:
$\begin{array}{l}
y = \dfrac{{\sqrt {x + 4} }}{{{x^2} - 5x + 6}} + \dfrac{1}{{\sqrt {5 - x} }}\\
Dkxd:\left\{ \begin{array}{l}
x + 4 \ge 0\\
5 - x > 0\\
{x^2} - 5x + 6 \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 4\\
x < 5\\
\left( {x - 2} \right)\left( {x - 3} \right) \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 4 \le x < 5\\
x \ne 2;x \ne 3
\end{array} \right.\\
Vậy\,TXD:D = \left[ { - 4;5} \right)\backslash \left\{ {2;3} \right\}
\end{array}$
