Phương pháp giải:

Sử dụng hằng đẳng thức \({a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right).\)

Lời giải chi tiết:

\(\begin{array}{l}A = {\left( {\frac{{1 + \sqrt 3 }}{2} - \frac{2}{{1 + \sqrt 3 }}} \right)^2} - {\left( {\frac{{1 - \sqrt 3 }}{2} - \frac{2}{{1 - \sqrt 3 }}} \right)^2}\\A = \left( {\frac{{1 + \sqrt 3 }}{2} - \frac{2}{{1 + \sqrt 3 }} + \frac{{1 - \sqrt 3 }}{2} - \frac{2}{{1 - \sqrt 3 }}} \right)\left( {\frac{{1 + \sqrt 3 }}{2} - \frac{2}{{1 + \sqrt 3 }} - \frac{{1 - \sqrt 3 }}{2} + \frac{2}{{1 - \sqrt 3 }}} \right)\\A = \left[ {\frac{{1 + \sqrt 3  + 1 - \sqrt 3 }}{2} - \left( {\frac{2}{{1 + \sqrt 3 }} + \frac{2}{{1 - \sqrt 3 }}} \right)} \right]\left[ {\frac{{1 + \sqrt 3  - 1 + \sqrt 3 }}{2} + \left( {\frac{2}{{1 - \sqrt 3 }} - \frac{2}{{1 + \sqrt 3 }}} \right)} \right]\\A = \left[ {1 - \frac{{2\left( {1 - \sqrt 3 } \right) + 2\left( {1 + \sqrt 3 } \right)}}{{1 - {{\left( {\sqrt 3 } \right)}^2}}}} \right].\left[ {\sqrt 3  + \frac{{2\left( {1 + \sqrt 3 } \right) - 2\left( {1 - \sqrt 3 } \right)}}{{1 - {{\left( {\sqrt 3 } \right)}^2}}}} \right]\\A = \left( {1 - \frac{{2 - 2\sqrt 3  + 2 + 2\sqrt 3 }}{{ - 2}}} \right)\left( {\sqrt 3  + \frac{{2 + 2\sqrt 3  - 2 + 2\sqrt 3 }}{{ - 2}}} \right)\\A = \left( {1 + \frac{4}{2}} \right)\left( {\sqrt 3  - \frac{{4\sqrt 3 }}{2}} \right) = 3.\frac{{ - 2\sqrt 3 }}{2}\\A =  - 3\sqrt 3 .\end{array}\)

Chọn B.