Đáp án:
$\begin{array}{l}
a)Dkxd:a > 0;a\# 1\\
A = \left( {\dfrac{{\sqrt a }}{{\sqrt a - 1}} - \dfrac{1}{{a - \sqrt a }}} \right):\left( {\dfrac{1}{{\sqrt a + 1}} + \dfrac{2}{{a - 1}}} \right)\\
= \dfrac{{\sqrt a .\sqrt a - 1}}{{\sqrt a \left( {\sqrt a - 1} \right)}}:\dfrac{{\sqrt a - 1 + 2}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{a - 1}}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}{{a + 1}}\\
= \dfrac{{a - 1}}{{\sqrt a }}.\dfrac{{\sqrt a + 1}}{{a + 1}}\\
= \dfrac{{\left( {a - 1} \right)\left( {\sqrt a + 1} \right)}}{{\sqrt a \left( {a + 1} \right)}}\\
b)a = 3 + 2\sqrt 2 \left( {tmdk} \right)\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\Leftrightarrow \sqrt a = \sqrt 2 + 1\\
\Leftrightarrow A = \dfrac{{\left( {3 + 2\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1 + 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {3 + 2\sqrt 2 + 1} \right)}}\\
= \dfrac{{\left( {2 + 2\sqrt 2 } \right)\left( {2 + \sqrt 2 } \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {4 + 2\sqrt 2 } \right)}}\\
= \dfrac{{2\left( {\sqrt 2 + 1} \right).\left( {2 + \sqrt 2 } \right)}}{{\left( {\sqrt 2 + 1} \right).2\left( {2 + \sqrt 2 } \right)}}\\
= 1\\
c)A < 0\\
\Leftrightarrow \dfrac{{\left( {a - 1} \right)\left( {\sqrt a + 1} \right)}}{{\sqrt a \left( {a + 1} \right)}} < 0\\
\Leftrightarrow a - 1 < 0\\
\Leftrightarrow a < 1\\
Vay\,0 < a < 1
\end{array}$
