Đáp án:$A=\sqrt{2}$
Giải thích các bước giải:
$\begin{array}{l}
A = \dfrac{{2 + \sqrt 3 }}{{\sqrt 2 + \sqrt {2 + \sqrt 3 } }} + \dfrac{{2 - \sqrt 3 }}{{\sqrt 2 - \sqrt {2 - \sqrt 3 } }}\\
= \dfrac{{\sqrt 2 \left( {2 + \sqrt 3 } \right)}}{{\sqrt 2 \left( {\sqrt 2 + \sqrt {2 + \sqrt 3 } } \right)}} + \dfrac{{\sqrt 2 \left( {2 - \sqrt 3 } \right)}}{{\sqrt 2 \left( {\sqrt 2 - \sqrt {2 - \sqrt 3 } } \right)}}\\
= \dfrac{{2\sqrt 2 + \sqrt 6 }}{{2 + \sqrt {4 + 2\sqrt 3 } }} + \dfrac{{2\sqrt 2 - \sqrt 6 }}{{2 - \sqrt {4 - 2\sqrt 3 } }}\\
= \dfrac{{2\sqrt 2 + \sqrt 6 }}{{2 + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} + \dfrac{{2\sqrt 2 - \sqrt 6 }}{{2 - \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}\\
= \dfrac{{2\sqrt 2 + \sqrt 6 }}{{2 + \sqrt 3 + 1}} + \dfrac{{2\sqrt 2 - \sqrt 6 }}{{2 - \left( {\sqrt 3 - 1} \right)}} = \dfrac{{2\sqrt 2 + \sqrt 6 }}{{3 + \sqrt 3 }} + \dfrac{{2\sqrt 2 - \sqrt 6 }}{{3 - \sqrt 3 }}\\
= \dfrac{{\sqrt 2 }}{{\sqrt 3 }}\left( {\dfrac{{2 + \sqrt 3 }}{{\sqrt 3 + 1}} + \dfrac{{2 - \sqrt 3 }}{{\sqrt 3 - 1}}} \right)\\
= \dfrac{{\sqrt 2 }}{{\sqrt 3 }}.\dfrac{{\left( {2 + \sqrt 3 } \right)\left( {\sqrt 3 - 1} \right) + \left( {2 - \sqrt 3 } \right)\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}\\
= \dfrac{{\sqrt 2 }}{{\sqrt 3 }}.\dfrac{{2\sqrt 3 }}{2} = \sqrt 2
\end{array}$
Vậy $A=\sqrt{2}$
