Chọn đáp án đúng.

Ta có $\dfrac{4}{{\sqrt 3  + 1}} + \dfrac{1}{{\sqrt 3  - 2}} + \dfrac{6}{{\sqrt 3  - 3}}$$= \dfrac{{4\left( {\sqrt 3  - 1} \right)}}{{\left( {\sqrt 3  - 1} \right)\left( {\sqrt 3  + 1} \right)}} + \dfrac{{1\left( {\sqrt 3  + 2} \right)}}{{\left( {\sqrt 3  + 2} \right)\left( {\sqrt 3  - 2} \right)}} + \dfrac{{6\left( {\sqrt 3  + 3} \right)}}{{\left( {\sqrt 3  + 3} \right)\left( {\sqrt 3  - 3} \right)}}$

$= \dfrac{{4\left( {\sqrt 3  - 1} \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - {1^2}}} + \dfrac{{\sqrt 3  + 2}}{{{{\left( {\sqrt 3 } \right)}^2} - {2^2}}} + \dfrac{{6\left( {\sqrt 3  + 3} \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - {3^2}}}$ $= \dfrac{{4\left( {\sqrt 3  - 1} \right)}}{{3 - 1}} + \dfrac{{\sqrt 3  + 2}}{{3 - 4}} + \dfrac{{6\left( {\sqrt 3  + 3} \right)}}{{3 - 9}}$

$= \dfrac{{4\left( {\sqrt 3  - 1} \right)}}{2} + \dfrac{{\sqrt 3  + 2}}{{\left( { - 1} \right)}} + \dfrac{{6\left( {\sqrt 3  + 3} \right)}}{{\left( { - 6} \right)}}$ $= 2\left( {\sqrt 3  - 1} \right) - \sqrt 3  - 2 - \sqrt 3  - 3 =  - 7.$