Lời giải
\(A = \frac{{1 + \frac{{\sqrt 3 }}{2}}}{{1 + \sqrt {1 + \frac{{\sqrt 3 }}{2}} }} + \frac{{1 - \frac{{\sqrt 3 }}{2}}}{{1 - \sqrt {1 - \frac{{\sqrt 3 }}{2}} }}\)
\( = \frac{{\frac{{2 + \sqrt 3 }}{2}}}{{1 + \sqrt {\frac{{4 + 2\sqrt 3 }}{4}} }} + \frac{{\frac{{2 - \sqrt 3 }}{2}}}{{1 - \sqrt {\frac{{4 - 2\sqrt 3 }}{4}} }}\)
\( = \frac{{\frac{{2 + \sqrt 3 }}{2}}}{{1 + \sqrt {\frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{4}} }} + \frac{{\frac{{2 - \sqrt 3 }}{2}}}{{1 - \sqrt {\frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{4}} }}\)
\( = \frac{{\frac{{2 + \sqrt 3 }}{2}}}{{1 + \frac{{\sqrt 3 + 1}}{2}}} + \frac{{\frac{{2 - \sqrt 3 }}{2}}}{{1 - \frac{{\sqrt 3 - 1}}{2}}}\)
\( = \frac{{\frac{{2 + \sqrt 3 }}{2}}}{{\frac{{3 + \sqrt 3 }}{2}}} + \frac{{\frac{{2 - \sqrt 3 }}{2}}}{{\frac{{3 - \sqrt 3 }}{2}}}\)
\( = \frac{{2 + \sqrt 3 }}{{3 + \sqrt 3 }} + \frac{{2 - \sqrt 3 }}{{3 - \sqrt 3 }}\)
\[ = \frac{{\left( {2 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right) + \left( {2 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}{{\left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)}}\]
\[ = \frac{{\left( {6 + \sqrt 3 - 3} \right) + \left( {6 - \sqrt 3 - 3} \right)}}{{9 - 3}}\]
\[ = \frac{6}{6} = 1\].
Vậy A = 1.