Gửi em!
\(\begin{array}{l} \dfrac{1}{{\sqrt 3 + 1}} + \dfrac{1}{{\sqrt 3 - 1}} - 2\sqrt 3 \\ = \dfrac{{1\left( {\sqrt 3 - 1} \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - {1^2}}} + \dfrac{{1\left( {\sqrt 3 + 1} \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - {1^2}}} - 2\sqrt 3 \\ = \dfrac{{\sqrt 3 - 1 + \sqrt 3 + 1 - 4\sqrt 3 }}{2}\\ = \dfrac{{ - 2\sqrt 3 }}{2} = - \sqrt 3 \end{array}\)