Đáp án:
$\begin{array}{l}
A = (\frac{{\sqrt a + 1}}{{\sqrt a - 1}} - \frac{{\sqrt a - 1}}{{\sqrt a + 1}} + 4\sqrt a )(\sqrt a + \frac{1}{{\sqrt a }})\\
A = \frac{{{{(\sqrt a + 1)}^2} - {{(\sqrt a - 1)}^2}}}{{a - 1}}.\frac{{a + 1}}{{\sqrt a }}\\
A = \frac{{4\sqrt a }}{{a - 1}}.\frac{{a + 1}}{{\sqrt a }} = \frac{{4(a + 1)}}{{a - 1}}\\
a = (4 + \sqrt {15} )(\sqrt {10} - \sqrt 6 )(\sqrt {4 - \sqrt {15} } )\\
2a = (8 + 2\sqrt {15} )(\sqrt 5 - \sqrt 3 )(\sqrt {8 - 2\sqrt {15} } )\\
2a = {(\sqrt 5 + \sqrt 3 )^2}(\sqrt 5 - \sqrt 3 )(\sqrt 5 - \sqrt 3 )\\
2a = 4\\
a = 2\\
A = 12\\
\end{array}$
