Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x\# \dfrac{1}{4};x\# 1\\
A = \left( {\dfrac{1}{{1 - \sqrt x }} - \dfrac{1}{{\sqrt x }}} \right)\\
:\left( {\dfrac{{2x + \sqrt x - 1}}{{1 - x}} + \dfrac{{2x\sqrt x + x - \sqrt x }}{{1 + x\sqrt x }}} \right)\\
= \dfrac{{\sqrt x - \left( {1 - \sqrt x } \right)}}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
:\left( {\dfrac{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} + \dfrac{{\sqrt x \left( {2x + \sqrt x - 1} \right)}}{{\left( {1 + \sqrt x } \right)\left( {x - \sqrt x + 1} \right)}}} \right)\\
= \dfrac{{2\sqrt x - 1}}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
:\left[ {\left( {2\sqrt x - 1} \right).\left( {\dfrac{1}{{1 - \sqrt x }} + \dfrac{{\sqrt x }}{{x - \sqrt x + 1}}} \right)} \right]\\
= \dfrac{{2\sqrt x - 1}}{{\sqrt x \left( {1 - \sqrt x } \right)}}:\left( {2\sqrt x - 1} \right)\\
:\left( {\dfrac{{x - \sqrt x + 1 + \sqrt x \left( {1 - \sqrt x } \right)}}{{\left( {1 - \sqrt x } \right)\left( {x - \sqrt x + 1} \right)}}} \right)\\
= \dfrac{1}{{\sqrt x \left( {1 - \sqrt x } \right)}}:\dfrac{1}{{\left( {1 - \sqrt x } \right)\left( {x - \sqrt x + 1} \right)}}\\
= \dfrac{1}{{\sqrt x \left( {1 - \sqrt x } \right)}}.\dfrac{{\left( {1 - \sqrt x } \right)\left( {x - \sqrt x + 1} \right)}}{1}\\
= \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
b)A = \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
= \sqrt x - 1 + \dfrac{1}{{\sqrt x }}\\
Theo\,Co - si:\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} = 2\\
\Leftrightarrow A \ge 2 - 1\\
\Leftrightarrow A \ge 1\\
\Leftrightarrow A \ge \sqrt A \\
Vậy\,A \ge \sqrt A
\end{array}$
