phần này hơi dài nên mik cho 50 điểm nha ^^

Đáp án:$\begin{array}{l}
a)\dfrac{{2a + 2\sqrt a  + 2}}{{\sqrt a }}\\
b)\,a = \dfrac{1}{4};a = 4\\
c)a > 0;a \ne 1
\end{array}$

Giải thích các bước giải:

$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
a > 0\\
a \ne 1
\end{array} \right.\\
a)A = \dfrac{{a\sqrt a  - 1}}{{a - \sqrt a }} - \dfrac{{a\sqrt a  + 1}}{{a + \sqrt a }}\\
 + \left( {\sqrt a  - \dfrac{1}{{\sqrt a }}} \right).\left( {\dfrac{{\sqrt a  + 1}}{{\sqrt a  - 1}} + \dfrac{{\sqrt a  - 1}}{{\sqrt a  + 1}}} \right)\\
 = \dfrac{{\left( {\sqrt a  - 1} \right)\left( {a + \sqrt a  + 1} \right)}}{{\sqrt a \left( {\sqrt a  - 1} \right)}} - \dfrac{{\left( {\sqrt a  + 1} \right)\left( {a - \sqrt a  + 1} \right)}}{{\sqrt a \left( {\sqrt a  + 1} \right)}}\\
 + \dfrac{{a - 1}}{{\sqrt a }}.\dfrac{{{{\left( {\sqrt a  + 1} \right)}^2} + {{\left( {\sqrt a  - 1} \right)}^2}}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}\\
 = \dfrac{{a + \sqrt a  + 1}}{{\sqrt a }} - \dfrac{{a - \sqrt a  + 1}}{{\sqrt a }}\\
 + \dfrac{{a - 1}}{{\sqrt a }}.\dfrac{{a + 2\sqrt a  + 1 + a - 2\sqrt a  + 1}}{{a - 1}}\\
 = \dfrac{{a + \sqrt a  + 1 - a + \sqrt a  - 1}}{{\sqrt a }} + \dfrac{1}{{\sqrt a }}.\dfrac{{2a + 2}}{1}\\
 = \dfrac{{2\sqrt a }}{{\sqrt a }} + \dfrac{{2a + 2}}{{\sqrt a }}\\
 = \dfrac{{2a + 2\sqrt a  + 2}}{{\sqrt a }}\\
b)A = 7\\
 \Leftrightarrow \dfrac{{2a + 2\sqrt a  + 2}}{{\sqrt a }} = 7\\
 \Leftrightarrow 2a + 2\sqrt a  + 2 = 7\sqrt a \\
 \Leftrightarrow 2a - 5\sqrt a  + 2 = 0\\
 \Leftrightarrow 2a - 4\sqrt a  - \sqrt a  + 2 = 0\\
 \Leftrightarrow \left( {\sqrt a  - 2} \right)\left( {2\sqrt a  - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt a  = 2\\
2\sqrt a  - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
a = 4\\
\sqrt a  = \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
a = 4\left( {tm} \right)\\
a = \dfrac{1}{4}\left( {tm} \right)
\end{array} \right.\\
Vay\,a = \dfrac{1}{4};a = 4\\
c)A > 6\\
 \Leftrightarrow \dfrac{{2a + 2\sqrt a  + 2}}{{\sqrt a }} > 6\\
 \Leftrightarrow \dfrac{{2a + 2\sqrt a  + 2}}{{\sqrt a }} - 6 > 0\\
 \Leftrightarrow \dfrac{{2a + 2\sqrt a  + 2 - 6\sqrt a }}{{\sqrt a }} > 0\\
 \Leftrightarrow 2a - 4\sqrt a  + 2 > 0\\
 \Leftrightarrow a - 2\sqrt a  + 1 > 0\\
 \Leftrightarrow {\left( {\sqrt a  - 1} \right)^2} > 0\\
 \Leftrightarrow \sqrt a  \ne 1\\
 \Leftrightarrow a \ne 1\\
Vậy\,a > 0;a \ne 1
\end{array}$