Lời giải chi tiết:

a) Hàm số xác định \(  \Leftrightarrow \left\{ \matrix{ x \ge 0 \hfill \cr  x - 1 \ge 0 \hfill \cr \sqrt x  + \sqrt {x - 1}  \ne 0 \hfill \cr  \sqrt x  - \sqrt {x - 1}  \ne 0 \hfill \cr 1 - \sqrt x  \ne 0 \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr x \ge 1 \hfill \cr \forall x \ge 1 \hfill \cr  \sqrt x  \ne \sqrt {x - 1}  \hfill \cr \sqrt x  \ne 1 \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{ x \ge 1 \hfill \cr x \ne x - 1 \hfill \cr  x \ne 1 \hfill \cr}  \right. \Leftrightarrow x > 1.  \)

b) Ta có:  \( \left( {\sqrt x  + \sqrt {x - 1} } \right)\left( {\sqrt x  - \sqrt {x - 1} } \right) = {\left( {\sqrt x } \right)^2} - {\left( {\sqrt {x - 1} } \right)^2} = x - \left( {x - 1} \right) = 1.\)

\( \eqalign{& C = {1 \over {\sqrt x  + \sqrt {x - 1} }} - {1 \over {\sqrt x  - \sqrt {x - 1} }} - {{x\sqrt x  - x} \over {1 - \sqrt x }}  \cr & \,\,\,\,\, = {{\sqrt x  - \sqrt {x - 1}  - \left( {\sqrt x  + \sqrt {x - 1} } \right)} \over {\left( {\sqrt x  + \sqrt {x - 1} } \right)\left( {\sqrt x  - \sqrt {x - 1} } \right)}} - {{x\left( {\sqrt x  - 1} \right)} \over {1 - \sqrt x }}  \cr & \,\,\,\,\, = {{\sqrt x  - \sqrt {x - 1}  - \sqrt x  - \sqrt {x - 1} } \over 1} + x\, = x - 2\sqrt {x - 1} . \cr} \)

Chọn D.