Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
A = \left( {\dfrac{1}{{\sqrt x - 2}} - \dfrac{1}{{\sqrt x + 2}}} \right).\dfrac{{\sqrt x - 2}}{2}\\
= \dfrac{{\left( {\sqrt x + 2} \right) - \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x - 2}}{2}\\
= \dfrac{4}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x - 2}}{2}\\
= \dfrac{2}{{\sqrt x + 2}}\\
b,\\
x = 9 \Rightarrow A = \dfrac{2}{{\sqrt 9 + 2}} = \dfrac{2}{5}\\
x = 9 - 4\sqrt 5 = 5 - 2.\sqrt 5 .2 + 4 = {\left( {\sqrt 5 - 2} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 5 - 2\\
\Rightarrow A = \dfrac{2}{{\left( {\sqrt 5 - 2} \right) + 2}} = \dfrac{2}{{\sqrt 5 }} = \dfrac{{2\sqrt 5 }}{5}\\
c,\\
A = \dfrac{3}{5} \Leftrightarrow \dfrac{2}{{\sqrt x + 2}} = \dfrac{3}{5}\\
\Leftrightarrow 2.5 = 3.\left( {\sqrt x + 2} \right)\\
\Leftrightarrow 10 = 3\sqrt x + 6\\
\Leftrightarrow \sqrt x = \dfrac{4}{3}\\
\Leftrightarrow x = \dfrac{{16}}{9}\\
d,\\
A > \dfrac{1}{2} \Leftrightarrow \dfrac{2}{{\sqrt x + 2}} > \dfrac{1}{2}\\
\Leftrightarrow 2.2 > 1.\left( {\sqrt x + 2} \right)\\
\Leftrightarrow 4 > \sqrt x + 2\\
\Leftrightarrow \sqrt x < 2\\
\Leftrightarrow 0 \le x < 4
\end{array}\)
