Đáp án:
$\begin{array}{l}
A = \dfrac{1}{{\sqrt 2 + 1}} - \sqrt {5 - 2\sqrt 6 } + \sqrt {12} \\
= \dfrac{{\sqrt 2 - 1}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} - \sqrt {3 - 2\sqrt 3 .\sqrt 2 + 2} + 2\sqrt 3 \\
= \dfrac{{\sqrt 2 - 1}}{{2 - 1}} - \sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} + 2\sqrt 3 \\
= \sqrt 2 - 1 - \left( {\sqrt 3 - \sqrt 2 } \right) + 2\sqrt 3 \\
= \sqrt 2 - 1 - \sqrt 3 + \sqrt 2 + 2\sqrt 3 \\
= \sqrt 3 + 2\sqrt 2 - 1\\
B = \sqrt {3 + 2\sqrt 2 } - \sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - \sqrt {\dfrac{{{{\left( {\sqrt 2 - 1} \right)}^2}}}{{2 - 1}}} \\
= \sqrt 2 + 1 - \left( {\sqrt 2 - 1} \right)\\
= 2\\
C = \sqrt {\dfrac{{2 - \sqrt 3 }}{2}} + \dfrac{{1 - \sqrt 3 }}{2}\\
= \dfrac{{\sqrt {4 - 2\sqrt 3 } }}{{\sqrt 4 }} + \dfrac{{1 - \sqrt 3 }}{2}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}{2} + \dfrac{{1 - \sqrt 3 }}{2}\\
= \dfrac{{\sqrt 3 - 1}}{2} + \dfrac{{1 - \sqrt 3 }}{2}\\
= 0\\
D = \left( {\sqrt {22} + 7\sqrt 2 } \right).\sqrt {30 - 7\sqrt {11} } \\
= \left( {\sqrt {11} + 7} \right).\sqrt 2 .\sqrt {30 - 7\sqrt {11} } \\
= \left( {\sqrt {11} + 7} \right).\sqrt {60 - 2.7.\sqrt {11} } \\
= \left( {\sqrt {11} + 7} \right).\sqrt {{{\left( {7 - \sqrt {11} } \right)}^2}} \\
= \left( {\sqrt {11} + 7} \right).\left( {7 - \sqrt {11} } \right)\\
= {7^2} - 11\\
= 49 - 11\\
= 38
\end{array}$
