Đáp án:
$\begin{array}{l}
1)a)A = \sqrt {6 - 4\sqrt 2 } - \sqrt {{{\left( {5 - 2\sqrt 2 } \right)}^2}} - 2\sqrt {\dfrac{1}{2}} \\
= \sqrt {4 - 2.2\sqrt 2 + 2} - \left( {5 - 2\sqrt 2 } \right) - 2.\dfrac{{\sqrt 2 }}{2}\\
= \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} - 5 + 2\sqrt 2 - \sqrt 2 \\
= 2 - \sqrt 2 - 5 + \sqrt 2 \\
= - 3\\
b)B = \sqrt {29 - 12\sqrt 5 } + \dfrac{8}{{\sqrt 5 - 1}} + \sqrt {20} \\
= \sqrt {20 - 2.2\sqrt 5 .3 + 9} + \dfrac{{8\left( {\sqrt 5 + 1} \right)}}{{5 - 1}} + 2\sqrt 5 \\
= \sqrt {{{\left( {2\sqrt 5 - 3} \right)}^2}} + 2\left( {\sqrt 5 + 1} \right) + 2\sqrt 5 \\
= 2\sqrt 5 - 3 + 2\sqrt 5 + 2 + 2\sqrt 5 \\
= 6\sqrt 5 - 1\\
c)C = \dfrac{{\sqrt 2 + 2}}{{\sqrt 2 + 1}} + \dfrac{2}{{\sqrt 2 - 1}} - \sqrt {12 - 8\sqrt 2 } \\
= \dfrac{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 + 1}} + \dfrac{{2\left( {\sqrt 2 + 1} \right)}}{{2 - 1}} - \sqrt {8 - 2.2\sqrt 2 .2 + 4} \\
= \sqrt 2 + 2\sqrt 2 + 2 - \sqrt {{{\left( {2\sqrt 2 - 2} \right)}^2}} \\
= 3\sqrt 2 + 2 - \left( {2\sqrt 2 - 2} \right)\\
= 4 + \sqrt 2 \\
2)a)A = \dfrac{3}{{\sqrt 7 - 2}} - \dfrac{3}{{\sqrt 7 + 2}}\\
= \dfrac{{3\left( {\sqrt 7 + 2} \right) - 3\left( {\sqrt 7 - 2} \right)}}{{\left( {\sqrt 7 - 2} \right)\left( {\sqrt 7 + 2} \right)}}\\
= \dfrac{{12}}{{7 - 4}}\\
= \dfrac{{12}}{3} = 4\\
b)\left\{ \begin{array}{l}
3\sqrt 5 = \sqrt {45} \\
2\sqrt 6 = \sqrt {24} \\
\sqrt {29} \\
4\sqrt 2 = \sqrt {32}
\end{array} \right.\\
\Leftrightarrow 2\sqrt 6 ;\sqrt {29} ;4\sqrt 2 ;3\sqrt 5
\end{array}$
