Đáp án:
$\begin{array}{l}
12)\sqrt {4 + \sqrt {10 + 2\sqrt 5 } } + \sqrt {4 - \sqrt {10 + 2\sqrt 5 } } = a\\
\Leftrightarrow {a^2} = 4 + \sqrt {10 + 2\sqrt 5 } + 4 - \sqrt {10 + 2\sqrt 5 } \\
+ 2.\sqrt {4 + \sqrt {10 + 2\sqrt 5 } } .\sqrt {4 - \sqrt {10 + 2\sqrt 5 } } \\
\Leftrightarrow {a^2} = 8 + 2.\sqrt {16 - \left( {10 + 2\sqrt 5 } \right)} \\
\Leftrightarrow {a^2} = 8 + 2.\sqrt {6 - 2\sqrt 5 } \\
\Leftrightarrow {a^2} = 8 + 2\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
\Leftrightarrow {a^2} = 8 + 2\left( {\sqrt 5 - 1} \right)\\
\Leftrightarrow {a^2} = 6 + 2\sqrt 5 = {\left( {\sqrt 5 + 1} \right)^2}\\
\Leftrightarrow a = \sqrt 5 + 1\\
Vay\,\sqrt {4 + \sqrt {10 + 2\sqrt 5 } } + \sqrt {4 - \sqrt {10 + 2\sqrt 5 } } = \sqrt 5 + 1\\
13)\\
\left( {5 + 2\sqrt 6 } \right)\left( {49 - 20\sqrt 6 } \right).\sqrt {5 - 2\sqrt 6 } \\
= {\left( {\sqrt 3 + \sqrt 2 } \right)^2}.\left( {49 - 20\sqrt 6 } \right).\sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \\
= \left( {\sqrt 3 + \sqrt 2 } \right).{\left( {5 - 2\sqrt 6 } \right)^2}.\left( {\sqrt 3 - \sqrt 2 } \right)\\
= \left( {\sqrt 3 + \sqrt 2 } \right)\left( {\sqrt 3 - \sqrt 2 } \right).\left( {49 - 20\sqrt 6 } \right)\\
= \left( {3 - 2} \right).\left( {49 - 20\sqrt 6 } \right)\\
= 49 - 20\sqrt 6 \\
14)\dfrac{1}{{\sqrt 2 + \sqrt {2 + \sqrt 3 } }} + \dfrac{1}{{\sqrt 2 - \sqrt {2 - \sqrt 3 } }}\\
= \dfrac{{\sqrt 2 }}{{2 + \sqrt {4 + 2\sqrt 3 } }} + \dfrac{{\sqrt 2 }}{{2 - \sqrt {4 - 2\sqrt 3 } }}\\
= \dfrac{{\sqrt 2 }}{{2 + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} + \dfrac{{\sqrt 2 }}{{2 - \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}\\
= \dfrac{{\sqrt 2 }}{{2 + \sqrt 3 + 1}} + \dfrac{{\sqrt 2 }}{{2 - \sqrt 3 + 1}}\\
= \dfrac{{\sqrt 2 }}{{3 + \sqrt 3 }} + \dfrac{{\sqrt 2 }}{{3 - \sqrt 3 }}\\
= \dfrac{{\sqrt 2 \left( {3 - \sqrt 3 } \right) + \sqrt 2 \left( {3 + \sqrt 3 } \right)}}{{\left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)}}\\
= \dfrac{{6\sqrt 2 }}{{9 - 3}}\\
= \sqrt 2
\end{array}$
