Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
2\sqrt x - 2 \ne 0\\
2\sqrt x + 2 \ne 0\\
1 - x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
b,\\
A = \dfrac{1}{{2\sqrt x - 2}} - \dfrac{1}{{2\sqrt x + 2}} + \dfrac{{\sqrt x }}{{1 - x}}\\
= \dfrac{1}{{2\left( {\sqrt x - 1} \right)}} - \dfrac{1}{{2\left( {\sqrt x + 1} \right)}} + \dfrac{{\sqrt x }}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right) - \left( {\sqrt x - 1} \right) - 2\sqrt x }}{{2.\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2 - 2\sqrt x }}{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 2.\left( {\sqrt x - 1} \right)}}{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 1}}{{\sqrt x + 1}}\\
x = \dfrac{4}{9} \Rightarrow A = \dfrac{{ - 1}}{{\sqrt {\dfrac{4}{9}} + 1}} = \dfrac{{ - 1}}{{\dfrac{2}{3} + 1}} = - \dfrac{3}{5}\\
c,\\
\left| A \right| = \dfrac{1}{3} \Leftrightarrow \left[ \begin{array}{l}
A = \dfrac{1}{3}\\
A = - \dfrac{1}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\dfrac{{ - 1}}{{\sqrt x + 1}} = \dfrac{1}{3}\\
\dfrac{{ - 1}}{{\sqrt x + 1}} = - \dfrac{1}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x + 1 = - 3\\
\sqrt x + 1 = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = - 4\\
\sqrt x = 2
\end{array} \right. \Leftrightarrow \sqrt x = 2 \Leftrightarrow x = 4
\end{array}\)
