Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{3 + \sqrt 5 }}{{\sqrt 2 + \sqrt {3 + \sqrt 5 } }} + \dfrac{{3 - \sqrt 5 }}{{\sqrt 2 - \sqrt {3 - \sqrt 5 } }}\\
= \dfrac{{\sqrt 2 .\left( {3 + \sqrt 5 } \right)}}{{\sqrt 2 .\left( {\sqrt 2 + \sqrt {3 + \sqrt 5 } } \right)}} + \dfrac{{\sqrt 2 .\left( {3 - \sqrt 5 } \right)}}{{\sqrt 2 .\left( {\sqrt 2 - \sqrt {3 - \sqrt 5 } } \right)}}\\
= \dfrac{{\sqrt 2 \left( {3 + \sqrt 5 } \right)}}{{2 + \sqrt {6 + 2\sqrt 5 } }} + \dfrac{{\sqrt 2 \left( {3 - \sqrt 5 } \right)}}{{2 - \sqrt {6 - 2\sqrt 5 } }}\\
= \dfrac{{\sqrt 2 \left( {3 + \sqrt 5 } \right)}}{{2 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }} + \dfrac{{\sqrt 2 .\left( {3 - \sqrt 5 } \right)}}{{2 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}\\
= \dfrac{{\sqrt 2 \left( {3 + \sqrt 5 } \right)}}{{2 + \sqrt 5 + 1}} + \dfrac{{\sqrt 2 \left( {3 - \sqrt 5 } \right)}}{{2 - \left( {\sqrt 5 - 1} \right)}}\\
= \dfrac{{\sqrt 2 \left( {3 + \sqrt 5 } \right)}}{{3 + \sqrt 5 }} + \dfrac{{\sqrt 2 \left( {3 - \sqrt 5 } \right)}}{{3 - \sqrt 5 }}\\
= \sqrt 2 + \sqrt 2 \\
= 2\sqrt 2
\end{array}\)
