Đáp án:
$\begin{array}{l}
a)A = \sqrt {4 + \sqrt {10 + 2\sqrt 5 } } + \sqrt {4 - \sqrt {10 + 2\sqrt 5 } } \\
\Rightarrow {A^2} = 4 + \sqrt {10 + 2\sqrt 5 } + \\
+ 2.\sqrt {4 + \sqrt {10 + 2\sqrt 5 } } .\sqrt {4 - \sqrt {10 + 2\sqrt 5 } } \\
+ 4 - \sqrt {10 + 2\sqrt 5 } \\
\Rightarrow {A^2} = 8 + 2.\sqrt {{4^2} - 10 - 2\sqrt 5 } \\
\Rightarrow {A^2} = 8 + 2\sqrt {6 - 2\sqrt 5 } \\
\Rightarrow {A^2} = 8 + 2.\left( {\sqrt 5 - 1} \right)\\
= 8 + 2\sqrt 5 - 2\\
= 6 + 2\sqrt 5 \\
= {\left( {\sqrt 5 + 1} \right)^2}\\
\Rightarrow A = \sqrt 5 + 1\left( {do:A > 0} \right)\\
b)\left( {5 + 2\sqrt 6 } \right)\left( {49 - 20\sqrt 6 } \right).\sqrt {5 - 2\sqrt 6 } \\
= {\left( {\sqrt 3 + \sqrt 2 } \right)^2}.\left( {25 - 2.2.\sqrt 6 .5 + 24} \right).\sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \\
= {\left( {\sqrt 3 + \sqrt 2 } \right)^2}.{\left( {5 - 2\sqrt 6 } \right)^2}.\left( {\sqrt 3 - \sqrt 2 } \right)\\
= \left( {\sqrt 3 + \sqrt 2 } \right).\left( {3 - 2} \right).{\left( {\sqrt 3 - \sqrt 2 } \right)^4}\\
= \left( {3 - 2} \right).{\left( {\sqrt 3 - \sqrt 2 } \right)^3}\\
= {\left( {\sqrt 3 - \sqrt 2 } \right)^3}\\
c)\dfrac{1}{{\sqrt 2 + \sqrt {2 + \sqrt 3 } }} + \dfrac{1}{{\sqrt 2 - \sqrt {2 - \sqrt 3 } }}\\
= \dfrac{{\sqrt 2 }}{{2 + \sqrt {4 + 2\sqrt 3 } }} + \dfrac{{\sqrt 2 }}{{2 - \sqrt {4 - 2\sqrt 3 } }}\\
= \dfrac{{\sqrt 2 }}{{2 + \sqrt 3 + 1}} + \dfrac{{\sqrt 2 }}{{2 - \left( {\sqrt 3 - 1} \right)}}\\
= \dfrac{{\sqrt 2 }}{{3 + \sqrt 3 }} + \dfrac{{\sqrt 2 }}{{3 - \sqrt 3 }}\\
= \dfrac{{\sqrt 2 \left( {3 - \sqrt 3 } \right) + \sqrt 2 \left( {3 + \sqrt 3 } \right)}}{{{3^2} - 3}}\\
= \dfrac{{6\sqrt 2 }}{6}\\
= \sqrt 2 \\
d)\sqrt {14 - 8\sqrt 3 } - \sqrt {24 - 12\sqrt 3 } \\
= \sqrt {8 - 2.2.2.\sqrt 3 + 6} - \sqrt {18 - 2.2.3.\sqrt 3 + 6} \\
= \sqrt {{{\left( {2\sqrt 2 - \sqrt 6 } \right)}^2}} - \sqrt {{{\left( {3\sqrt 2 - \sqrt 6 } \right)}^2}} \\
= 2\sqrt 2 - \sqrt 6 - 3\sqrt 2 + \sqrt 6 \\
= - \sqrt 2
\end{array}$
