Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 1;x \ne 3\\
A = \dfrac{1}{{\sqrt x - \sqrt {x - 1} }} - \dfrac{{x - 3}}{{\sqrt {x - 1} - \sqrt 2 }}\\
= \dfrac{{\sqrt x + \sqrt {x - 1} }}{{\left( {\sqrt x - \sqrt {x - 1} } \right)\left( {\sqrt x + \sqrt {x - 1} } \right)}}\\
- \dfrac{{\left( {x - 3} \right)\left( {\sqrt {x - 1} + \sqrt 2 } \right)}}{{\left( {\sqrt {x - 1} + \sqrt 2 } \right)\left( {\sqrt {x - 1} - \sqrt 2 } \right)}}\\
= \dfrac{{\sqrt x + \sqrt {x - 1} }}{{x - \left( {x - 1} \right)}} - \dfrac{{\left( {x - 3} \right)\left( {\sqrt {x - 1} + \sqrt 2 } \right)}}{{x - 1 - 2}}\\
= \dfrac{{\sqrt x + \sqrt {x - 1} }}{1} - \dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt {x - 1} + \sqrt 2 } \right)}}{{x - 3}}\\
= \sqrt x + \sqrt {x - 1} - \left( {\sqrt {x - 1} + \sqrt 2 } \right)\\
= \sqrt x - \sqrt 2 \\
b)x = 3 - 2\sqrt 2 \left( {tmdk} \right)\\
\Rightarrow x = 2 - 2\sqrt 2 + 1\\
\Rightarrow x = {\left( {\sqrt 2 - 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 2 - 1\\
\Rightarrow A = \sqrt x - \sqrt 2 = \sqrt 2 - 1 - \sqrt 2 = - 1
\end{array}$
