Đáp án:
$\begin{array}{l}
a\sqrt {8 + 4\sqrt 3 } - \sqrt {8 - 4\sqrt 3 } \\
= \sqrt {2.\left( {4 + 2\sqrt 3 } \right)} - \sqrt {2\left( {4 - 2\sqrt 3 } \right)} \\
= \sqrt 2 .\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \sqrt 2 .\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
= \sqrt 2 .\left( {\sqrt 3 + 1} \right) - \sqrt 2 \left( {\sqrt 3 - 1} \right)\\
= \sqrt 6 + \sqrt 2 - \sqrt 6 + \sqrt 2 \\
= 2\sqrt 2 \\
b)\sqrt {9 - 4\sqrt 5 } - \sqrt {9 + 4\sqrt 5 } \\
= \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 + 2} \right)}^2}} \\
= \sqrt 5 - 2 - \left( {\sqrt 5 + 2} \right)\\
= - 4\\
c)\sqrt {21 + 8\sqrt 5 } - \sqrt {21 - 8\sqrt 5 } \\
= \sqrt {{{\left( {4 + \sqrt 5 } \right)}^2}} - \sqrt {{{\left( {4 - \sqrt 5 } \right)}^2}} \\
= 4 + \sqrt 5 - \left( {4 - \sqrt 5 } \right)\\
= 2\sqrt 5 \\
a)\dfrac{{\sqrt {7 + 4\sqrt 3 } }}{{\sqrt 3 + 2}} = \dfrac{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} }}{{\sqrt 3 + 2}} = \dfrac{{2 + \sqrt 3 }}{{\sqrt 3 + 2}} = 1\\
b)\dfrac{{\sqrt {9 - 4\sqrt 5 } }}{{2 - \sqrt 5 }} = \dfrac{{\sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} }}{{2 - \sqrt 5 }} = \dfrac{{\sqrt 5 - 2}}{{2 - \sqrt 5 }} = - 1\\
c)\dfrac{{\sqrt {7 - 4\sqrt 3 } }}{{\sqrt {2 - \sqrt 3 } }}.\sqrt {2 + \sqrt 3 } \\
= \dfrac{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} }}{{\sqrt {2 - \sqrt 3 } }}.\sqrt {2 + \sqrt 3 } \\
= \dfrac{{2 - \sqrt 3 }}{{\sqrt {2 - \sqrt 3 } }}.\sqrt {2 + \sqrt 3 } \\
= \sqrt {2 - \sqrt 3 } .\sqrt {2 + \sqrt 3 } \\
= \sqrt {{2^2} - 3} \\
= 1
\end{array}$
