Đáp án:
$\begin{array}{l}
d)\sqrt {2 - \sqrt 3 } .\left( {\sqrt 6 + \sqrt 2 } \right)\\
= \sqrt {2 - \sqrt 3 } .\sqrt 2 .\left( {\sqrt 3 + 1} \right)\\
= \sqrt {4 - 2\sqrt 3 } .\left( {\sqrt 3 + 1} \right)\\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} .\left( {\sqrt 3 + 1} \right)\\
= \left( {\sqrt 3 - 1} \right).\left( {\sqrt 3 + 1} \right)\\
= 3 - 1\\
= 2\\
e)\sqrt {3 - \sqrt 5 } + \sqrt {3 + \sqrt 5 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 2 .\sqrt {3 - \sqrt 5 } + \sqrt 2 .\sqrt {3 + \sqrt 5 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {6 - 2\sqrt 5 } + \sqrt {6 + 2\sqrt 5 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 5 - 1 + \sqrt 5 + 1} \right)\\
= \dfrac{1}{{\sqrt 2 }}.2\sqrt 5 \\
= \sqrt 2 .\sqrt 5 \\
= \sqrt {10}
\end{array}$
