Đáp án:
$\begin{array}{l}
a)\sqrt {{{\left( {\sqrt 2 + 3} \right)}^2}} .\sqrt {11 - 6\sqrt 2 } \\
= \left( {\sqrt 2 + 3} \right).\sqrt {9 - 2.3.\sqrt 2 + 2} \\
= \left( {\sqrt 2 + 3} \right).\sqrt {{{\left( {3 - \sqrt 2 } \right)}^2}} \\
= \left( {3 + \sqrt 2 } \right).\left( {3 - \sqrt 2 } \right)\\
= 9 - 2\\
= 7\\
b)\sqrt {{{\left( {\sqrt 5 - 3} \right)}^2}} .\sqrt {\dfrac{1}{{3 - \sqrt 5 }}} \\
= \left( {3 - \sqrt 5 } \right).\sqrt {\dfrac{{3 + \sqrt 5 }}{{9 - 5}}} \\
= \left( {3 - \sqrt 5 } \right).\dfrac{{\sqrt {3 + \sqrt 5 } }}{{\sqrt 4 }}\\
= \sqrt {3 - \sqrt 5 } .\sqrt {\left( {3 - \sqrt 5 } \right).\left( {3 + \sqrt 5 } \right)} .\dfrac{1}{2}\\
= \dfrac{1}{{2\sqrt 2 }}.\sqrt {6 - 2\sqrt 5 } .\sqrt {9 - 5} \\
= \dfrac{1}{{2\sqrt 2 }}.\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} .2\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 5 - 1} \right)\\
= \dfrac{{\sqrt {10} - \sqrt 2 }}{2}\\
c)\left( {\sqrt 6 - 2\sqrt 3 + 5\sqrt 2 - \dfrac{1}{2}\sqrt 8 } \right).2\sqrt 6 \\
= 12 - 12\sqrt 2 + 20\sqrt 3 - 4\sqrt 3 \\
= 12 - 12\sqrt 2 + 16\sqrt 3 \\
d)\left( {\sqrt {18} + 2\sqrt {75} - 3\sqrt {24} } \right).3\sqrt 2 \\
= \left( {3\sqrt 2 + 10\sqrt 3 - 6\sqrt 6 } \right).3\sqrt 2 \\
= 18 + 30\sqrt 6 - 36\sqrt 3 \\
e) - 3.\sqrt {\dfrac{{{{\left( {a - b} \right)}^3}.{b^5}}}{c}} .\dfrac{2}{3}.\sqrt {\dfrac{{{c^3}}}{{3\left( {a - b} \right)}}} .\sqrt {48b} \\
= - 2.\dfrac{{\left( {a - b} \right).{b^2}\sqrt {a - b} }}{{\sqrt c }}.\dfrac{{c\sqrt c }}{{\sqrt 3 .\sqrt {a - b} }}.4\sqrt {3b} \\
= - 8.\left( {a - b} \right).{b^2}\sqrt b .c\\
f)\left( {\sqrt {ab} + 3\sqrt {\dfrac{a}{b}} - 5\sqrt {\dfrac{b}{a}} + 2\sqrt {\dfrac{1}{{ab}}} } \right).\sqrt {ab} \\
= ab + 3a - 5b + 2
\end{array}$
