Đáp án:
$\begin{array}{l}
a)Dkxd:a > 0;a\# 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 }}{1}.\dfrac{1}{{\sqrt a }}\\
= 4a\\
b)a = \left( {2 + \sqrt 3 } \right)\left( {\sqrt 3 - 1} \right)\sqrt {2 - \sqrt 3 } \\
= \dfrac{1}{{2\sqrt 2 }}.\left( {4 + 2\sqrt 3 } \right)\left( {\sqrt 3 - 1} \right)\sqrt {4 - 2\sqrt 3 } \\
= \dfrac{1}{{2\sqrt 2 }}.{\left( {\sqrt 3 + 1} \right)^2}\left( {\sqrt 3 - 1} \right)\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
= \dfrac{1}{{2\sqrt 2 }}.{\left( {\sqrt 3 + 1} \right)^2}\left( {\sqrt 3 - 1} \right).\left( {\sqrt 3 - 1} \right)\\
= \dfrac{1}{{2\sqrt 2 }}.{\left( {3 - 1} \right)^2}\\
= \dfrac{1}{{2\sqrt 2 }}{.2^2}\\
= \sqrt 2 \left( {tmdk} \right)\\
\Leftrightarrow A = 4a = 4\sqrt 2 \\
c)\dfrac{A}{{4\left( {\sqrt a - 1} \right)}} > 1\\
\Leftrightarrow \dfrac{{4a}}{{4\left( {\sqrt a - 1} \right)}} > 1\\
\Leftrightarrow \dfrac{{a - \sqrt a + 1}}{{\sqrt a - 1}} > 0\\
\Leftrightarrow \sqrt a - 1 > 0\\
\Leftrightarrow \sqrt a > 1\\
\Leftrightarrow a > 1\\
Vậy\,a > 1\\
d)M = \dfrac{A}{{\sqrt a - 1}} = \dfrac{{4a}}{{\sqrt a - 1}}\\
= \dfrac{{4a - 4 + 4}}{{\sqrt a - 1}}\\
= 4.\left( {\sqrt a + 1 + \dfrac{1}{{\sqrt a - 1}}} \right)\\
= 4.\left( {\sqrt a - 1 + \dfrac{1}{{\sqrt a - 1}} + 2} \right)\\
Do:\sqrt a - 1 + \dfrac{1}{{\sqrt a - 1}} \ge 2\sqrt {\left( {\sqrt a - 1} \right).\dfrac{1}{{\sqrt a - 1}}} = 2\\
\Leftrightarrow \sqrt a - 1 + \dfrac{1}{{\sqrt a - 1}} + 2 \ge 4\\
\Leftrightarrow 4\left( {\sqrt a - 1 + \dfrac{1}{{\sqrt a - 1}} + 2} \right) \ge 16\\
\Leftrightarrow M \ge 16\\
\Leftrightarrow GTNN:M = 16\,\\
Khi:\sqrt a - 1 = \dfrac{1}{{\sqrt a - 1}} \Leftrightarrow \sqrt a - 1 = 1 \Leftrightarrow a = 4\left( {tmdk} \right)
\end{array}$
