Đáp án:

\(\begin{array}{l}
a,\\
A = \dfrac{{a + \sqrt a  + 1}}{{\sqrt a  - 1}}\\
b,\\
a > 1.\\
c,\\
A = \dfrac{{2007 - \sqrt {2006} }}{{2006 - 2\sqrt {2006} }}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
DKXD:\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
a + 1 \ne 0\\
\sqrt a  - 1 \ne 0\\
a\sqrt a  + \sqrt a  - a - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a \ge 0\\
a \ne 1
\end{array} \right.\\
a,\\
A = \left( {1 + \dfrac{{\sqrt a }}{{a + 1}}} \right):\left( {\dfrac{1}{{\sqrt a  - 1}} - \dfrac{{2\sqrt a }}{{a\sqrt a  + \sqrt a  - a - 1}}} \right)\\
 = \dfrac{{\left( {a + 1} \right) + \sqrt a }}{{a + 1}}:\left( {\dfrac{1}{{\sqrt a  - 1}} - \dfrac{{2\sqrt a }}{{\sqrt a .\left( {a + 1} \right) - \left( {a + 1} \right)}}} \right)\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}:\left( {\dfrac{1}{{\sqrt a  - 1}} - \dfrac{{2\sqrt a }}{{\left( {a + 1} \right)\left( {\sqrt a  - 1} \right)}}} \right)\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}:\dfrac{{\left( {a + 1} \right) - 2\sqrt a }}{{\left( {\sqrt a  - 1} \right)\left( {a + 1} \right)}}\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}:\dfrac{{a - 2\sqrt a  + 1}}{{\left( {\sqrt a  - 1} \right)\left( {a + 1} \right)}}\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}:\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{\left( {\sqrt a  - 1} \right)\left( {a + 1} \right)}}\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}:\dfrac{{\sqrt a  - 1}}{{a + 1}}\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}.\dfrac{{a + 1}}{{\sqrt a  - 1}}\\
 = \dfrac{{a + \sqrt a  + 1}}{{\sqrt a  - 1}}\\
b,\\
A > 1 \Leftrightarrow \dfrac{{a + \sqrt a  + 1}}{{\sqrt a  - 1}} > 1\\
 \Leftrightarrow \dfrac{{a + \sqrt a  + 1}}{{\sqrt a  - 1}} - 1 > 0\\
 \Leftrightarrow \dfrac{{\left( {a + \sqrt a  + 1} \right) - \left( {\sqrt a  - 1} \right)}}{{\sqrt a  - 1}} > 0\\
 \Leftrightarrow \dfrac{{a + \sqrt a  + 1 - \sqrt a  + 1}}{{\sqrt a  - 1}} > 0\\
 \Leftrightarrow \dfrac{{a + 2}}{{\sqrt a  - 1}} > 0\\
a \ge 0,a \ne 0 \Rightarrow a + 2 > 0\\
 \Rightarrow \sqrt a  - 1 > 0\\
 \Leftrightarrow \sqrt a  > 1\\
 \Leftrightarrow a > 1.\\
c,\\
a = 2007 - 2\sqrt {2006}  = 2006 - 2\sqrt {2006} .1 + 1 = {\left( {\sqrt {2006}  - 1} \right)^2}\\
 \Rightarrow \sqrt a  = \sqrt {{{\left( {\sqrt {2006}  - 1} \right)}^2}}  = \left| {\sqrt {2006}  - 1} \right| = \sqrt {2006}  - 1\\
 \Rightarrow A = \dfrac{{a + \sqrt a  + 1}}{{\sqrt a  - 1}}\\
 = \dfrac{{2007 - 2\sqrt {2006}  + \sqrt {2006}  - 1 + 1}}{{2007 - 2\sqrt {2006}  - 1}}\\
 = \dfrac{{2007 - \sqrt {2006} }}{{2006 - 2\sqrt {2006} }}
\end{array}\)