Đáp án:

\(\begin{array}{l}
a)\dfrac{{2a + 2\sqrt a  + 2}}{{\sqrt a }}\\
b)\left[ \begin{array}{l}
a = 4\\
a = \dfrac{1}{4}
\end{array} \right.\\
c)a \ne 1;a > 0\\
d)Min = 2\sqrt 2  + 2
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)N = \dfrac{{\left( {\sqrt a  - 1} \right)\left( {a + \sqrt a  + 1} \right)}}{{\sqrt a \left( {\sqrt a  - 1} \right)}} - \dfrac{{\left( {\sqrt a  + 1} \right)\left( {a - \sqrt a  + 1} \right)}}{{\sqrt a \left( {\sqrt a  + 1} \right)}} + \dfrac{{a - 1}}{{\sqrt a }}.\dfrac{{a + 2\sqrt a  + 1 + a - 2\sqrt a  + 1}}{{a - 1}}\\
 = \dfrac{{a + \sqrt a  + 1 - \left( {a - \sqrt a  + 1} \right)}}{{\sqrt a }} + \dfrac{{a - 1}}{{\sqrt a }}.\dfrac{{2a + 2}}{{a - 1}}\\
 = \dfrac{{2\sqrt a }}{{\sqrt a }} + \dfrac{{2a + 2}}{{\sqrt a }}\\
 = \dfrac{{2a + 2\sqrt a  + 2}}{{\sqrt a }}\\
b)N = 7\\
 \to \dfrac{{2a + 2\sqrt a  + 2}}{{\sqrt a }} = 7\\
 \to 2a + 2\sqrt a  + 2 = 7\sqrt a \\
 \to 2a - 5\sqrt a  + 2 = 0\\
 \to \left( {\sqrt a  - 2} \right)\left( {2\sqrt a  - 1} \right) = 0\\
 \to \left[ \begin{array}{l}
\sqrt a  - 2 = 0\\
2\sqrt a  - 1 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
a = 4\\
a = \dfrac{1}{4}
\end{array} \right.\\
c)N > 6\\
 \to \dfrac{{2a + 2\sqrt a  + 2}}{{\sqrt a }} > 6\\
 \to \dfrac{{2a + 2\sqrt a  + 2 - 6\sqrt a }}{{\sqrt a }} > 0\\
 \to 2a - 4\sqrt a  + 2 > 0\left( {do:\sqrt a  > 0\forall a > 0} \right)\\
 \to a - 2\sqrt a  + 1 > 0\\
 \to {\left( {\sqrt a  - 1} \right)^2} > 0\\
 \to \sqrt a  - 1 \ne 0\\
 \to a \ne 1;a > 0\\
d)N - \sqrt a  = \dfrac{{2a + 2\sqrt a  + 2}}{{\sqrt a }} - \sqrt a \\
 = \dfrac{{2a + 2\sqrt a  + 2 - a}}{{\sqrt a }}\\
 = \dfrac{{a + 2\sqrt a  + 2}}{{\sqrt a }}\\
 = \sqrt a  + 2 + \dfrac{2}{{\sqrt a }}\\
Do:a > 0\\
BDT:Co - si:\sqrt a  + \dfrac{2}{{\sqrt a }} \ge 2\sqrt {\sqrt a .\dfrac{2}{{\sqrt a }}}  = 2\sqrt 2 \\
 \to \sqrt a  + \dfrac{2}{{\sqrt a }} + 2 \ge 2\sqrt 2  + 2\\
 \to Min = 2\sqrt 2  + 2\\
 \Leftrightarrow \sqrt a  = \dfrac{2}{{\sqrt a }}\\
 \to a = 2
\end{array}\)