Lời giải chi tiết:

\(  \eqalign{ & N = \left( {{1 \over {\sqrt x  - 3}} - {1 \over {\sqrt x }}} \right):\left( {{{\sqrt x  + 3} \over {\sqrt x  - 2}} - {{\sqrt x  + 2} \over {\sqrt x  - 3}}} \right)  \cr & \,\,\,\,\, = {{\sqrt x  - \sqrt x  + 3} \over {\sqrt x \left( {\sqrt x  - 3} \right)}}:{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right) - \left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)} \over {\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}}  \cr  & \,\,\,\,\,\, = {3 \over {\sqrt x \left( {\sqrt x  - 3} \right)}}:{{x - 9 - x + 4} \over {\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}}  \cr  & \,\,\,\,\,\, = {3 \over {\sqrt x \left( {\sqrt x  - 3} \right)}}.{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)} \over { - 5}}\, = {{ - 3\left( {\sqrt x  - 2} \right)} \over {5\sqrt x }} = {{6 - 3\sqrt x } \over {5\sqrt x }}. \cr} \)

Chọn D.