Đáp án:
\[ = \dfrac{2}{{\sqrt x - 2}}\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\left( {\dfrac{1}{{\sqrt x - 2}} - \dfrac{1}{{\sqrt x + 2}}} \right).\dfrac{{\sqrt x + 2}}{2}\\
= \dfrac{{\left( {\sqrt x + 2} \right) - \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right).\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 2}}{2}\\
= \dfrac{{\sqrt x + 2 - \sqrt x + 2}}{{\left( {\sqrt x - 2} \right).\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 2}}{2}\\
= \dfrac{4}{{\left( {\sqrt x - 2} \right).\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 2}}{2}\\
= \dfrac{2}{{\sqrt x - 2}}
\end{array}\)
