Đáp án:
$\begin{array}{l}
Dkxd:a > 0;a\# 1\\
1)A = \left( {\dfrac{{\sqrt a + 1}}{{\sqrt a - 1}} - \dfrac{{\sqrt a - 1}}{{\sqrt a + 1}} + 4\sqrt a } \right).\left( {\sqrt a - \dfrac{1}{{\sqrt a }}} \right)\\
= \dfrac{{{{\left( {\sqrt a + 1} \right)}^2} - {{\left( {\sqrt a - 1} \right)}^2} + 4\sqrt a \left( {a - 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}.\dfrac{{a - 1}}{{\sqrt a }}\\
= \dfrac{{a + 2\sqrt a + 1 - a + 2\sqrt a - 1 + 4a\sqrt a - 4\sqrt a }}{{a - 1}}.\dfrac{{a - 1}}{{\sqrt a }}\\
= \dfrac{{4a\sqrt a }}{{\sqrt a }}\\
= 4a\\
2)a = \dfrac{{\sqrt 6 }}{{2 + \sqrt 6 }}\left( {tmdk} \right)\\
A = 4a = 4.\dfrac{{\sqrt 6 }}{{2 + \sqrt 6 }}\\
= \dfrac{{4\sqrt 6 \left( {\sqrt 6 - 2} \right)}}{{6 - 4}}\\
= 2\sqrt 6 \left( {\sqrt 6 - 2} \right)\\
= 12 - 4\sqrt 6 \\
3)\sqrt A > A\\
\Leftrightarrow \sqrt A .\left( {\sqrt A - 1} \right) < 0\\
\Leftrightarrow \sqrt A - 1 < 0\left( {do:\sqrt A > 0} \right)\\
\Leftrightarrow \sqrt A < 1\\
\Leftrightarrow A < 1\\
\Leftrightarrow 4a < 1\\
\Leftrightarrow a < \dfrac{1}{4}\\
Vậy\,0 < a < \dfrac{1}{4}
\end{array}$
