Đáp án:
$\begin{array}{l}
1)\dfrac{{\sqrt {\sqrt 5 + \sqrt 2 } }}{{\sqrt {3\sqrt 5 - 3\sqrt 2 } }} = \dfrac{{\sqrt {\sqrt 5 + \sqrt 2 } .\sqrt {\sqrt 5 + \sqrt 2 } }}{{\sqrt 3 .\sqrt {\sqrt 5 - \sqrt 2 } .\sqrt {\sqrt 5 + \sqrt 2 } }}\\
= \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 3 .\sqrt {\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)} }}\\
= \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 3 .\sqrt {5 - 2} }}\\
= \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 3 .\sqrt 3 }}\\
= \dfrac{{\sqrt 5 + \sqrt 2 }}{3}\\
2)B = \sqrt {3 + \sqrt {5 - 2\sqrt 3 } } - \sqrt {3 - \sqrt {5 - 2\sqrt 3 } } \\
\Leftrightarrow {B^2} = 3 + \sqrt {5 - 2\sqrt 3 } - 2.\sqrt {3 + \sqrt {5 - 2\sqrt 3 } } .\sqrt {3 - \sqrt {5 - 2\sqrt 3 } } \\
+ 3 - \sqrt {5 - 2\sqrt 3 } \\
= 6 - 2.\sqrt {{3^2} - \left( {5 - 2\sqrt 3 } \right)} \\
= 6 - 2\sqrt {4 + 2\sqrt 3 } \\
= 6 - 2\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= 6 - 2.\left( {\sqrt 3 + 1} \right)\\
= 4 - 2\sqrt 3 \\
= {\left( {\sqrt 3 - 1} \right)^2}\\
\Leftrightarrow B = \sqrt 3 - 1\left( {do:B > 0} \right)\\
3)\\
\dfrac{{\sqrt {3 - \sqrt 5 } }}{{\sqrt {\sqrt 5 - 1} }}.\sqrt {2 + 2\sqrt 5 } \\
= \dfrac{{\sqrt {6 - 2\sqrt 5 } }}{{\sqrt {\sqrt 5 - 1} }}.\sqrt {\sqrt 5 + 1} \\
= \dfrac{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}{{\sqrt {\sqrt 5 - 1} }}.\sqrt {\sqrt 5 + 1} \\
= \dfrac{{\sqrt 5 - 1}}{{\sqrt {\sqrt 5 - 1} }}.\sqrt {\sqrt 5 + 1} \\
= \sqrt {\sqrt 5 - 1} .\sqrt {\sqrt 5 + 1} \\
= \sqrt {\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)} \\
= \sqrt {5 - 1} \\
= \sqrt 4 = 2\\
4)\sqrt {\dfrac{2}{{8 + 3\sqrt 7 }}} + \sqrt {\dfrac{{38 - 14\sqrt 7 }}{{3 - \sqrt 7 }}} \\
= \sqrt {\dfrac{4}{{16 + 2.3.\sqrt 7 }}} + \sqrt {\dfrac{{\left( {38 - 14\sqrt 7 } \right)\left( {3 + \sqrt 7 } \right)}}{{9 - 7}}} \\
= \sqrt {\dfrac{4}{{{{\left( {3 + \sqrt 7 } \right)}^2}}}} + \sqrt {\dfrac{{16 - 4\sqrt 7 }}{2}} \\
= \dfrac{2}{{3 + \sqrt 7 }} + \sqrt {8 - 2\sqrt 7 } \\
= \dfrac{{2\left( {3 - \sqrt 7 } \right)}}{{{3^2} - 7}} + \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} \\
= \dfrac{{2\left( {3 - \sqrt 7 } \right)}}{2} + \sqrt 7 - 1\\
= 3 - \sqrt 7 + \sqrt 7 - 1\\
= 2
\end{array}$
