Lời giải chi tiết:

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

\( \eqalign{& b)\,A = {{2\sqrt x } \over {\sqrt x  + 3}} - {{\sqrt x  - 1} \over {3 - \sqrt x }} - {{3 - 11\sqrt x } \over {x - 9}}  \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {{2\sqrt x } \over {\sqrt x  + 3}} + {{\sqrt x  - 1} \over {\sqrt x  - 3}} - {{3 - 11\sqrt x } \over {\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}  \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {{2\sqrt x \left( {\sqrt x  - 3} \right) + \left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right) - 3 + 11\sqrt x } \over {\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}  \cr  & \,\,\,\,\,\,\,\,\,\,\,\, = {{2x - 6\sqrt x  + x + 2\sqrt x  - 3 - 3 + 11\sqrt x } \over {\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}  \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \,{{3x + 7\sqrt x  - 6} \over {\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}  \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {{\left( {3\sqrt x  - 2} \right)\left( {\sqrt x  + 3} \right)} \over {\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}} = {{3\sqrt x  - 2} \over {\sqrt x  - 3}}. \cr}  \)

Chọn C.