Đáp án:
$\begin{array}{l}
a)2\sqrt 5 - \sqrt {125} - \sqrt {80} + \sqrt {605} \\
= 2\sqrt 5 - 5\sqrt 5 - 4\sqrt 5 + 11\sqrt 5 \\
= 4\sqrt 5 \\
b)\sqrt {15 - \sqrt {216} } + \sqrt {33 - 12\sqrt 6 } \\
= \sqrt {15 - 6\sqrt 6 } + \sqrt {33 - 12\sqrt 6 } \\
= \sqrt {9 - 2.3.\sqrt 6 + 6} + \sqrt {24 - 2.2\sqrt 6 .3 + 9} \\
= \sqrt {{{\left( {3 - \sqrt 6 } \right)}^2}} + \sqrt {{{\left( {2\sqrt 6 - 3} \right)}^2}} \\
= 3 - \sqrt 6 + 2\sqrt 6 - 3\\
= \sqrt 6 \\
c)\sqrt {8\sqrt 3 } - 2\sqrt {25\sqrt {12} } + 4\sqrt {\sqrt {192} } \\
= 2\sqrt {2.\sqrt 3 } - 2.5\sqrt {2\sqrt 3 } + 4.\sqrt {4.2\sqrt 3 } \\
= 2\sqrt {2\sqrt 3 } - 10\sqrt {2\sqrt 3 } + 8\sqrt {2\sqrt 3 } \\
= 0\\
d)\sqrt {2 - \sqrt 3 } .\left( {\sqrt 6 + \sqrt 2 } \right)\\
= \sqrt {2 - \sqrt 3 } .\sqrt 2 .\left( {\sqrt 3 + 1} \right)\\
= \sqrt {4 - 2\sqrt 3 } .\left( {\sqrt 3 + 1} \right)\\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} .\left( {\sqrt 3 + 1} \right)\\
= \left( {\sqrt 3 - 1} \right).\left( {\sqrt 3 + 1} \right)\\
= 3 - 1\\
= 2\\
e)\sqrt {3 - \sqrt 5 } + \sqrt {3 + \sqrt 5 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {6 - 2\sqrt 5 } + \sqrt {6 + 2\sqrt 5 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 5 - 1 + \sqrt 5 + 1} \right)\\
= \dfrac{1}{{\sqrt 2 }}.2\sqrt 5 \\
= \sqrt {10}
\end{array}$
