Đáp án:
$\begin{array}{l}
Dkxd:x > 0;x \ne 3;x \ne 2\\
\left( {\dfrac{1}{{\sqrt x - \sqrt {x - 1} }} - \dfrac{{x - 3}}{{\sqrt {x - 1} - \sqrt 2 }}} \right).\\
\left( {\dfrac{2}{{\sqrt 2 - \sqrt x }} - \dfrac{{\sqrt x + \sqrt 2 }}{{\sqrt {2x} - x}}} \right)\\
= \left( {\dfrac{{\sqrt x + \sqrt {x - 1} }}{{x - \left( {x - 1} \right)}} - \dfrac{{\left( {x - 3} \right)\left( {\sqrt {x - 1} + \sqrt 2 } \right)}}{{x - 1 - 2}}} \right)\\
.\dfrac{{2.\sqrt x - \left( {\sqrt x + \sqrt 2 } \right)}}{{\sqrt x .\left( {\sqrt 2 - \sqrt x } \right)}}\\
= \left( {\dfrac{{\sqrt x + \sqrt {x - 1} }}{1} - \dfrac{{\left( {x - 3} \right)\left( {\sqrt {x - 1} + \sqrt 2 } \right)}}{{x - 3}}} \right)\\
.\dfrac{{\sqrt x - \sqrt 2 }}{{\sqrt x .\left( {\sqrt 2 - \sqrt x } \right)}}\\
= \left( {\sqrt x + \sqrt {x - 1} - \left( {\sqrt {x - 1} + \sqrt 2 } \right)} \right).\dfrac{{ - 1}}{{\sqrt x }}\\
= \left( {\sqrt x - \sqrt 2 } \right).\dfrac{{ - 1}}{{\sqrt x }}\\
= \dfrac{{\sqrt 2 - \sqrt x }}{{\sqrt x }}
\end{array}$
