Đáp án + Giải thích các bước giải:

$\textbf{2}\bigg)$

$\begin{array}{l} \textbf{A} &=\dfrac{1}{\sqrt{2} + \sqrt{2 + \sqrt{3}}} + \dfrac{1}{\sqrt{2} - \sqrt{2 - \sqrt{3}}}
\\ &= \dfrac{\sqrt{2} - \sqrt{2 + \sqrt{3}}}{\left(\sqrt{2} + \sqrt{2 + \sqrt{3}}\right)\left(\sqrt{2} - \sqrt{2 + \sqrt{3}}\right)} + \dfrac{\sqrt{2} + \sqrt{2 - \sqrt{3}}}{\left(\sqrt{2} - \sqrt{2 - \sqrt{3}}\right)\left(\sqrt{2} + \sqrt{2 - \sqrt{3}}\right)}
\\ &= \dfrac{\sqrt{2} - \sqrt{2 + \sqrt{3}}}{2 - (2 + \sqrt{3})} + \dfrac{\sqrt{2} + \sqrt{2 - \sqrt{3}}}{2 - (2 - \sqrt{3})}
\\ &= \dfrac{\sqrt{2} - \sqrt{2 + \sqrt{3}}}{-\sqrt{3}} + \dfrac{\sqrt{2} + \sqrt{2 - \sqrt{3}}}{\sqrt{3}}
\\ &= \dfrac{\sqrt{2} + \sqrt{2 - \sqrt{3}} - \left(\sqrt{2} - \sqrt{2 + \sqrt{3}}\right)}{\sqrt{3}}
\\ &= \dfrac{\sqrt{2 - \sqrt{3}} + \sqrt{2 + \sqrt{3}}}{\sqrt{3}}
\\ &= \dfrac{\sqrt{4 - 2\sqrt{3}} + \sqrt{4 + 2\sqrt{3}}}{\sqrt{6}}
\\ &= \dfrac{\sqrt{(\sqrt{3} - 1)^2} + \sqrt{(\sqrt{3} + 1)^2}}{\sqrt{6}}
\\ &= \dfrac{\sqrt{3} - 1 + \sqrt{3} + 1}{\sqrt{6}}
\\ &= \sqrt{2}\end{array}$

$\begin{array}{l} \textbf{A} &=\dfrac{3 + \sqrt{5}}{\sqrt{2}+ \sqrt{3 + \sqrt{5}}} + \dfrac{3 - \sqrt{5}}{\sqrt{2} - \sqrt{3 - \sqrt{5}}}
\\ &= \dfrac{6 + 2\sqrt{5}}{2 + \sqrt{6 + 2\sqrt{5}}} + \dfrac{6 - 2\sqrt{5}}{2 - \sqrt{6 - 2\sqrt{5}}}
\\ &= \dfrac{\sqrt{2}(3 + \sqrt{5})}{2 + \sqrt{(\sqrt{5} + 1)^2}}+ \dfrac{\sqrt{2}(3 - \sqrt{5})}{2 - \sqrt{(\sqrt{5} - 1)^2}}
\\ &= \dfrac{\sqrt{2}(3 + \sqrt{5})}{2 + \sqrt{5} + 1}+ \dfrac{\sqrt{2}(3 - \sqrt{5})}{2 - \sqrt{5} + 1}
\\ &= \dfrac{\sqrt{2}(3 + \sqrt{5})}{3 + \sqrt{5}}+ \dfrac{\sqrt{2}(3 - \sqrt{5})}{3 - \sqrt{5}}
\\ &= \sqrt{2} + \sqrt{2}
\\ &= 2\sqrt{2}\end{array}$