Đáp án:
$\begin{array}{l}
19)\sqrt {\dfrac{{3 + 2\sqrt 3 }}{{4 + 2\sqrt 3 }}} = \sqrt {\dfrac{{3 + 2\sqrt 3 }}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}} = \dfrac{{\sqrt {3 + 2\sqrt 3 } }}{{\sqrt 3 + 1}}\\
= \dfrac{{\sqrt {3 + 2\sqrt 3 } .\left( {\sqrt 3 - 1} \right)}}{2}\\
20)\sqrt {\dfrac{{29 + 4\sqrt 7 }}{{5 + 2\sqrt 6 }}} = \sqrt {\dfrac{{28 + 2.2\sqrt 7 + 1}}{{3 + 2\sqrt 3 .\sqrt 2 + 2}}} \\
= \sqrt {\dfrac{{{{\left( {2\sqrt 7 + 1} \right)}^2}}}{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}} = \dfrac{{2\sqrt 7 + 1}}{{\sqrt 3 + \sqrt 2 }} = \dfrac{{\left( {2\sqrt 7 + 1} \right)\left( {\sqrt 3 - \sqrt 2 } \right)}}{{3 - 2}}\\
= 2\sqrt {21} - 2\sqrt {14} + \sqrt 3 - \sqrt 2 \\
21)\sqrt {\dfrac{{11 + 2\sqrt {30} }}{{3 + 2\sqrt 2 }}} = \sqrt {\dfrac{{6 + 2\sqrt 6 .\sqrt 5 + 5}}{{2 + 2\sqrt 2 + 1}}} \\
= \sqrt {\dfrac{{{{\left( {\sqrt 6 + \sqrt 5 } \right)}^2}}}{{{{\left( {\sqrt 2 + 1} \right)}^2}}}} = \dfrac{{\sqrt 6 + \sqrt 5 }}{{\sqrt 2 + 1}}\\
= \dfrac{{\left( {\sqrt 6 + \sqrt 5 } \right)\left( {\sqrt 2 - 1} \right)}}{{2 - 1}}\\
= 2\sqrt 3 - \sqrt 6 + \sqrt {10} - \sqrt 5
\end{array}$
