Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x \ne 1\\
A = \left( {\dfrac{1}{{1 - \sqrt x }} + \dfrac{1}{{1 + \sqrt x }}} \right):\left( {\dfrac{1}{{1 - \sqrt x }} - \dfrac{1}{{1 + \sqrt x }}} \right)\\
+ \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{{1 + \sqrt x + 1 - \sqrt x }}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}:\dfrac{{1 + \sqrt x - 1 + \sqrt x }}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
+ \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{2}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}.\dfrac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{2\sqrt x }}\\
+ \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{1}{{\sqrt x }} + \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{{1 - \sqrt x + \sqrt x }}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
= \dfrac{1}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
b)x > 0;x \ne 1\\
x = 7 + 4\sqrt 3 \left( {tmdk} \right)\\
= 4 + 2.2.\sqrt 3 + 3\\
= {\left( {2 + \sqrt 3 } \right)^2}\\
\Rightarrow \sqrt x = 2 + \sqrt 3 \\
\Rightarrow A = \dfrac{1}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
= \dfrac{1}{{\left( {2 + \sqrt 3 } \right)\left( {1 - 2 - \sqrt 3 } \right)}}\\
= \dfrac{1}{{\left( {2 + \sqrt 3 } \right)\left( { - 1 - \sqrt 3 } \right)}}\\
= \dfrac{{2 - \sqrt 3 }}{{ - 1 - \sqrt 3 }}\\
= \dfrac{{\left( {\sqrt 3 - 2} \right)\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}\\
= \dfrac{{3 - \sqrt 3 - 2\sqrt 3 + 2}}{{3 - 2}}\\
= 5 - 3\sqrt 3 \\
c)A = \dfrac{1}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
Do:\sqrt x \left( {1 - \sqrt x } \right)\\
= - x + \sqrt x \\
= - \left( {x - \sqrt x } \right)\\
= - \left( {x - 2.\dfrac{1}{2}.\sqrt x + \dfrac{1}{4}} \right) + \dfrac{1}{4}\\
= - {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\
\Rightarrow \sqrt x \left( {1 - \sqrt x } \right) \le \dfrac{1}{4}\\
\Rightarrow \dfrac{1}{{\sqrt x \left( {1 - \sqrt x } \right)}} \ge 4\\
\Rightarrow A \ge 4\\
\Rightarrow GTNN:A = 4\,\\
Khi:\sqrt x - \dfrac{1}{2} = 0 \Rightarrow x = \dfrac{1}{4}\left( {tmdk} \right)
\end{array}$
