Lời giải chi tiết:

Ta có

\(\begin{array}{l}A = \frac{{\sqrt 2 }}{{2\sqrt 2 + \sqrt {3 + \sqrt 5 } }} + \frac{{\sqrt 2 }}{{2\sqrt 2 - \sqrt {3 - \sqrt 5 } }}\\= \frac{{\sqrt 2 .\sqrt 2 }}{{2\sqrt 2 .\sqrt 2 + \sqrt 2 .\sqrt {3 + \sqrt 5 } }} + \frac{{\sqrt 2 .\sqrt 2 }}{{2\sqrt 2 .\sqrt 2 - \sqrt 2 .\sqrt {3 - \sqrt 5 } }}\\= \frac{2}{{4 + \sqrt {6 + 2\sqrt 5 } }} + \frac{2}{{4 - \sqrt {6 - 2\sqrt 5 } }}\\= \frac{2}{{4 + \sqrt {{{\sqrt 5 }^2} + 2\sqrt 5 + 1} }} + \frac{2}{{4 - \sqrt {{{\sqrt 5 }^2} - 2\sqrt 5 + 1} }}\\= \frac{2}{{4 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }} + \frac{2}{{4 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}\\= \frac{2}{{4 + \sqrt 5 + 1}} + \frac{2}{{4 - \sqrt 5 + 1}}\\= \frac{2}{{5 + \sqrt 5 }} + \frac{2}{{5 - \sqrt 5 }}\\= \frac{{2.\left( {5 - \sqrt 5 } \right) + 2.\left( {5 + \sqrt 5 } \right)}}{{\left( {5 + \sqrt 5 } \right).\left( {5 - \sqrt 5 } \right)}}\\= \frac{{10 - 2\sqrt 5 + 10 + 2\sqrt 5 }}{{20}} = 1\end{array}\)