Lời giải chi tiết:

a) Hàm số xác định \(  \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr \sqrt x  - 1 \ne 0 \hfill \cr  \sqrt x  - 3 \ne 0 \hfill \cr 1 - {1 \over {\sqrt x  - 1}} \ne 0 \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr \sqrt x  \ne 1 \hfill \cr \sqrt x  \ne 3 \hfill \cr  {1 \over {\sqrt x  - 1}} \ne 1 \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr x \ne 1 \hfill \cr x \ne 9 \hfill \cr \sqrt x  - 1 \ne 1 \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr x \ne 1 \hfill \cr x \ne 9 \hfill \cr  \sqrt x  \ne 2 \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr  x \ne 1 \hfill \cr x \ne 4 \hfill \cr x \ne 9 \hfill \cr}  \right..  \)

\( \eqalign{& b)\,B = \left( {{{\sqrt x  + 2} \over {\sqrt x  - 1}} - {{\sqrt x  + 1} \over {\sqrt x  - 3}} + {{3\sqrt x  - 1} \over {\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 3} \right)}}} \right):\left( {1 - {1 \over {\sqrt x  - 1}}} \right)  \cr  & \,\,\,\,\,\,\,\,\,\,\, = {{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 3} \right) - \left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right) + 3\sqrt x  - 1} \over {\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 3} \right)}}:{{\sqrt x  - 1 - 1} \over {\sqrt x  - 1}}  \cr & \,\,\,\,\,\,\,\,\,\,\, = {{x - \sqrt x  - 6 - x + 1 + 3\sqrt x  - 1} \over {\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 3} \right)}}.{{\sqrt x  - 1} \over {\sqrt x  - 2}}  \cr & \,\,\,\,\,\,\,\,\,\,\, = {{2\sqrt x  - 6} \over {\left( {\sqrt x  - 3} \right)\left( {\sqrt x  - 2} \right)}} = {{2\left( {\sqrt x  - 3} \right)} \over {\left( {\sqrt x  - 3} \right)\left( {\sqrt x  - 2} \right)}} = {2 \over {\sqrt x  - 2}}. \cr}   \)

Chọn D.