Điều kiện xác định:
$\left\{ \begin{array}{l} \sqrt A + 2 \ne 0\\ \sqrt A - 2 \ne 0\\ A \ge 0\\ \sqrt A \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} A > 0\\ A \ne 4 \end{array} \right.$
$\begin{array}{l} \left( {\dfrac{1}{{\sqrt A + 2}} + \dfrac{1}{{\sqrt A - 2}}} \right)\dfrac{{\sqrt A - 2}}{{\sqrt A }}\\ = \dfrac{{\sqrt A - 2 + \sqrt A + 2}}{{\left( {\sqrt A - 2} \right)\left( {\sqrt A + 2} \right)}}.\dfrac{{\sqrt A - 2}}{{\sqrt A }}\\ = \dfrac{2}{{\sqrt A + 2}} \end{array}$
