Giải thích các bước giải:
$\begin{array}{l}
a)2\sqrt 2 \left( {\sqrt 3 - 2} \right) + {\left( {1 + 2\sqrt 2 } \right)^2} - 2\sqrt 6 \\
= 2\sqrt 6 - 4\sqrt 2 + 9 + 4\sqrt 2 - 2\sqrt 6 \\
= 9\\
b)\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } \\
= \dfrac{1}{{\sqrt 2 }}\left( {\sqrt {4 + 2\sqrt 3 } + \sqrt {4 - 2\sqrt 3 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}\left( {\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} } \right)\\
= \dfrac{1}{{\sqrt 2 }}\left( {\left| {\sqrt 3 + 1} \right| + \left| {\sqrt 3 - 1} \right|} \right)\\
= \dfrac{1}{{\sqrt 2 }}\left( {\sqrt 3 + 1 + \sqrt 3 - 1} \right)\\
= \dfrac{{2\sqrt 3 }}{{\sqrt 2 }}\\
= \sqrt 6 \\
c)\sqrt {\dfrac{4}{{{{\left( {2 - \sqrt 5 } \right)}^2}}}} - \sqrt {\dfrac{4}{{{{\left( {2 + \sqrt 5 } \right)}^2}}}} \\
= \sqrt {{{\left( {\dfrac{2}{{2 - \sqrt 5 }}} \right)}^2}} - \sqrt {{{\left( {\dfrac{2}{{2 + \sqrt 5 }}} \right)}^2}} \\
= \left| {\dfrac{2}{{2 - \sqrt 5 }}} \right| - \left| {\dfrac{2}{{2 + \sqrt 5 }}} \right|\\
= \dfrac{2}{{\sqrt 5 - 2}} - \dfrac{2}{{\sqrt 5 + 2}}\\
= 2\left( {\dfrac{1}{{\sqrt 5 - 2}} - \dfrac{1}{{\sqrt 5 + 2}}} \right)\\
= 2\left( {\dfrac{{\sqrt 5 + 2 - \left( {\sqrt 5 - 2} \right)}}{{{{\left( {\sqrt 5 } \right)}^2} - {2^2}}}} \right)\\
= 2.\dfrac{4}{{5 - 4}}\\
= 8
\end{array}$
Ta có điều phải chứng minh
