Đáp án:
$\begin{array}{l}
a)\left( {3 + \sqrt 5 } \right)\left( {\sqrt {10} - 2} \right)\left( {\sqrt {3 - \sqrt 5 } } \right)\\
= \left( {3 + \sqrt 5 } \right)\left( {\sqrt 5 - \sqrt 2 } \right).\sqrt 2 .\sqrt {3 - \sqrt 5 } \\
= \left( {3 + \sqrt 5 } \right).\left( {\sqrt 5 - \sqrt 2 } \right).\sqrt {6 - 2\sqrt 5 } \\
= \dfrac{1}{2}.\left( {6 + 2\sqrt 5 } \right).\left( {\sqrt 5 - \sqrt 2 } \right).\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \dfrac{1}{2}{\left( {\sqrt 5 + 1} \right)^2}.\left( {\sqrt 5 - 1} \right).\left( {\sqrt 5 - \sqrt 2 } \right)\\
= \dfrac{1}{2}.\left( {\sqrt 5 + 1} \right).\left( {5 - 1} \right).\left( {\sqrt 5 - 2} \right)\\
= 2.\left( {\sqrt 5 + 1} \right)\left( {\sqrt 5 - 2} \right)\\
= 6 - 2\sqrt 5 \\
b)\left( {4 + \sqrt {15} } \right)\left( {\sqrt {10} - \sqrt 6 } \right).\sqrt {4 - \sqrt {15} } \\
= \dfrac{1}{2}.\left( {8 + 2\sqrt {15} } \right).\left( {\sqrt 5 - \sqrt 3 } \right).\sqrt 2 .\sqrt {4 - \sqrt {15} } \\
= \dfrac{1}{2}\left( {5 + 2\sqrt 5 .\sqrt 3 + 3} \right).\left( {\sqrt 5 - \sqrt 3 } \right).\sqrt {8 - 2\sqrt {15} } \\
= \dfrac{1}{2}.{\left( {\sqrt 5 + \sqrt 3 } \right)^2}\left( {\sqrt 5 - \sqrt 3 } \right).\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \\
= \dfrac{1}{2}.{\left( {\sqrt 5 + \sqrt 3 } \right)^2}.\left( {\sqrt 5 - \sqrt 3 } \right).\left( {\sqrt 5 - \sqrt 3 } \right)\\
= \dfrac{1}{2}.{\left( {\sqrt 5 + \sqrt 3 } \right)^2}.{\left( {\sqrt 5 - \sqrt 3 } \right)^2}\\
= \dfrac{1}{2}.{\left( {5 - 3} \right)^2}\\
= 2
\end{array}$
