Câu c)
$\begin{array}{l} a = \sqrt {4 + \sqrt {10 + 2\sqrt 5 } } + \sqrt {4 - \sqrt {10 + 2\sqrt 5 } } \\ \Rightarrow {a^2} = 4 + \sqrt {10 + 2\sqrt 5 } + 4 - \sqrt {10 + 2\sqrt 5 } + 2.\sqrt {\left( {4 + \sqrt {10 + 2\sqrt 5 } } \right)\left( {4 - \sqrt {10 + 2\sqrt 5 } } \right)} \\ \Rightarrow {a^2} = 8 + 2.\sqrt {16 - \left( {10 + 2\sqrt 5 } \right)} \\ \Rightarrow {a^2} = 8 + 2.\sqrt {6 - 2\sqrt 5 } = 8 + 2\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} = 8 + 2\left( {\sqrt 5 - 1} \right) = 6 + 2\sqrt 5 = {\left( {\sqrt 5 + 1} \right)^2}\\ \Rightarrow a = \sqrt 5 + 1\left( {a > 0} \right)\\ \Rightarrow \sqrt {4 + \sqrt {10 + 2\sqrt 5 } } + \sqrt {4 - \sqrt {10 + 2\sqrt 5 } } = \sqrt 5 + 1 \end{array}$
