Đáp án đúng: C
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.