$A = \sqrt {x + 2} .\sqrt {x - 3} $ có nghĩa thì :
$\begin{array}{l} \left\{ \begin{array}{l} x + 2 \ge 0\\ x - 3 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge - 2\\ x \ge 3 \end{array} \right.\\ \Leftrightarrow x \ge 3 \end{array}$
$B = \sqrt {\left( {x + 2} \right)\left( {x - 3} \right)}$ có nghĩa thì:
$\begin{array}{l} \left( {x + 2} \right)\left( {x - 3} \right) \ge 0\\ \Leftrightarrow \left[ \begin{array}{l} x \le - 2\\ x \ge 3 \end{array} \right. \end{array}$
$\Rightarrow \begin{array}{l} AB = \left( {\sqrt {x + 2} .\sqrt {x - 3} } \right).\sqrt {\left( {x + 2} \right)\left( {x - 3} \right)} \\ \left\{ \begin{array}{l} x \ge - 2\\ x \ge 3\\ \left[ \begin{array}{l} x \ge 3\\ x \le - 2 \end{array} \right. \end{array} \right. \Rightarrow x \ge 3 \end{array}$
$D = \left[ {3; + \infty } \right)$
