Đáp án:
$\begin{array}{l}
a){\left( {\sqrt {2 - \sqrt 3 } - \sqrt {2 + \sqrt 3 } } \right)^2}\\
= 2 - \sqrt 3 - 2.\sqrt {2 - \sqrt 3 } .\sqrt {2 + \sqrt 3 } + 2 + \sqrt 3 \\
= 4 - 2\sqrt {4 - 3} \\
= 4 - 2\\
= 2\\
b)\left( {1 + \sqrt 3 + \sqrt 5 } \right)\left( {1 + \sqrt 3 - \sqrt 5 } \right)\\
= {\left( {1 + \sqrt 3 } \right)^2} - 5\\
= 1 + 2\sqrt 3 + 3 - 5\\
= 2\sqrt 3 - 1\\
c)\left( {1 + \sqrt 2 + \sqrt 3 } \right)\left( {1 - \sqrt 2 - \sqrt 3 } \right)\\
= 1 - {\left( {\sqrt 2 + \sqrt 3 } \right)^2}\\
= 1 - \left( {5 + 2\sqrt 2 .\sqrt 3 } \right)\\
= - 4 - 2\sqrt 6 \\
d)\sqrt {3 + \sqrt 5 } - \sqrt {3 - \sqrt 5 } - \sqrt 2 \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {6 + 2\sqrt 5 } - \sqrt {6 - 2\sqrt 5 } - 2} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} - 2} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 5 + 1 - \sqrt 5 + 1 - 2} \right)\\
= 0\\
e)\sqrt {\sqrt 3 - \sqrt 2 } .\sqrt {\sqrt 3 + \sqrt 2 } \\
= \sqrt {3 - 2} \\
= 1\\
f)F = \sqrt {4 + \sqrt 7 } + \sqrt {4 - \sqrt 7 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {8 + 2\sqrt 7 } + \sqrt {8 - 2\sqrt 7 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 7 + 1 + \sqrt 7 - 1} \right)\\
= \dfrac{1}{{\sqrt 2 }}.2\sqrt 7 \\
= \sqrt 2 .\sqrt 7 \\
= \sqrt {14}
\end{array}$
