Lời giải chi tiết:

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

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

Chọn B.