Đáp án:
\(\dfrac{{3\sqrt x - 6}}{{x\sqrt x - 3\sqrt x }}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:\left\{ \begin{array}{l}
x \ge 1\\
\sqrt {x - 1} \ne \sqrt 2 \\
x \ne 2
\end{array} \right. \to \left\{ \begin{array}{l}
x \ge 1\\
x \ne 3\\
x \ne 2
\end{array} \right.\\
\dfrac{{\sqrt x + \sqrt {x - 1} }}{{x - \left( {x - 1} \right)}} - \dfrac{{\left( {x - 3} \right)\left( {\sqrt {x - 1} + \sqrt 2 } \right)}}{{x - 1 - 2}}\\
= \dfrac{{\sqrt x + \sqrt {x - 1} }}{1} - \dfrac{{x\sqrt {x - 1} + x\sqrt x - 3\sqrt {x - 1} - 3\sqrt 2 }}{{x - 3}}\\
= \dfrac{{x\sqrt x + x\sqrt {x - 1} - 3\sqrt x - 3\sqrt {x - 1} - x\sqrt {x - 1} - x\sqrt x + 3\sqrt {x - 1} + 3\sqrt 2 }}{{x - 3}}\\
= \dfrac{{ - 3\sqrt x + 3\sqrt 2 }}{{x - 3}}\\
\dfrac{2}{{\sqrt 2 - \sqrt x }} - \dfrac{{\sqrt x + 2}}{{\sqrt x \left( {\sqrt 2 - \sqrt x } \right)}}\\
= \dfrac{{2\sqrt x - \sqrt x - 2}}{{\sqrt x \left( {\sqrt 2 - \sqrt x } \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x \left( {\sqrt 2 - \sqrt x } \right)}}\\
\left( {\dfrac{1}{{\sqrt x - \sqrt {x - 1} }} - \dfrac{{x - 3}}{{\sqrt {x - 1} - \sqrt 2 }}} \right).\left( {\dfrac{2}{{\sqrt 2 - \sqrt x }} - \dfrac{{\sqrt x + 2}}{{\sqrt {2x} - x}}} \right)\\
= \dfrac{{ - 3\sqrt x + 3\sqrt 2 }}{{x - 3}}.\dfrac{{\sqrt x - 2}}{{\sqrt x \left( {\sqrt 2 - \sqrt x } \right)}}\\
= \dfrac{{3\left( {\sqrt 2 - \sqrt x } \right)}}{{x - 3}}.\dfrac{{\sqrt x - 2}}{{\sqrt x \left( {\sqrt 2 - \sqrt x } \right)}}\\
= \dfrac{{3\left( {\sqrt x - 2} \right)}}{{x\sqrt x - 3\sqrt x }} = \dfrac{{3\sqrt x - 6}}{{x\sqrt x - 3\sqrt x }}
\end{array}\)
