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