Đáp án:
$\begin{array}{l}
a)\left( {\sqrt {27} - 3\sqrt 2 + 2\sqrt 6 } \right):3\sqrt 3 \\
= \left( {3\sqrt 3 - 3\sqrt 2 + 2\sqrt 6 } \right):3\sqrt 3 \\
= \dfrac{{3\sqrt 3 }}{{3\sqrt 3 }} - \dfrac{{3\sqrt 2 }}{{3\sqrt 3 }} + \dfrac{{2\sqrt 6 }}{{3\sqrt 3 }}\\
= 1 - \dfrac{{\sqrt 6 }}{3} + \dfrac{{2\sqrt 2 }}{3}\\
= \dfrac{{3 - \sqrt 6 + 2\sqrt 2 }}{3}\\
b)\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + \sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} \\
= \sqrt 3 + 1 + \sqrt 3 - 1\\
= 2\sqrt 3 \\
c){\left( {\sqrt 2 + 1} \right)^2} + {\left( {\sqrt 2 - 1} \right)^2}\\
= 2 + 2\sqrt 2 + 1 + 2 - 2\sqrt 2 + 1\\
= 6\\
d){\left( {\sqrt 3 + 1} \right)^2} + {\left( {1 - \sqrt 3 } \right)^2}\\
= 3 + 2\sqrt 3 + 1 + 1 - 2\sqrt 3 + 3\\
= 8\\
e)\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - \sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} \\
= \sqrt 2 + 1 - \left( {\sqrt 2 - 1} \right)\\
= 2\\
f)\sqrt {7 + 4\sqrt 3 } + \sqrt {7 - 4\sqrt 3 } \\
= \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} \\
= 2 + \sqrt 3 - 2 + \sqrt 3 \\
= 2\sqrt 3 \\
g)\sqrt {6 + 2\sqrt 5 } + \sqrt {6 - 2\sqrt 5 } \\
= \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \sqrt 5 + 1 + \sqrt 5 - 1\\
= 2\sqrt 5 \\
h)\sqrt {4 - \sqrt 7 } - \sqrt {4 + \sqrt 7 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {8 - 2\sqrt 7 } - \sqrt {8 + 2\sqrt 7 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 7 - 1 - \sqrt 7 - 1} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( { - 2} \right)\\
= \sqrt 2 \\
i)\left( {3 - \sqrt 5 } \right)\left( {\sqrt {10} - \sqrt 2 } \right)\sqrt {3 + \sqrt 5 } \\
= \left( {3 - \sqrt 5 } \right)\left( {\sqrt 5 - 1} \right).\sqrt 2 .\sqrt {3 + \sqrt 5 } \\
= \left( {3 - \sqrt 5 } \right).\left( {\sqrt 5 - 1} \right).\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} \\
= \left( {3 - \sqrt 5 } \right).\left( {\sqrt 5 - 1} \right).\left( {\sqrt 5 + 1} \right)\\
= \left( {3 - \sqrt 5 } \right).\left( {5 - 1} \right)\\
= 4\left( {3 - \sqrt 5 } \right)\\
= 12 - 4\sqrt 5
\end{array}$
