Đáp án:
$\begin{array}{l}
1)b)\sqrt {3{{\left( {2 - \sqrt 5 } \right)}^2}} \\
= \sqrt 3 .\left| {2 - \sqrt 5 } \right|\\
= \sqrt 3 .\left( {\sqrt 5 - 2} \right)\\
= \sqrt {15} - 2\sqrt 3 \\
e)\sqrt {9 - 4\sqrt 5 } - \sqrt {9 + 4\sqrt 5 } \\
= \sqrt {5 - 2.2.\sqrt 5 + 4} - \sqrt {5 + 2.2.\sqrt 5 + 4} \\
= \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 + 2} \right)}^2}} \\
= \sqrt 5 - 2 - \left( {\sqrt 5 + 2} \right)\\
= \sqrt 5 - 2 - \sqrt 5 - 2\\
= - 4\\
f)\sqrt {{{\left( {\sqrt 3 + 2} \right)}^2}} - \sqrt {{{\left( {\sqrt 3 - 2} \right)}^2}} \\
= \sqrt 3 + 2 - \left( {2 - \sqrt 3 } \right)\\
= 2\sqrt 3 \\
2)c)\dfrac{1}{{\sqrt 5 + \sqrt 3 }} - \dfrac{1}{{\sqrt 5 - \sqrt 3 }}\\
= \dfrac{{\sqrt 5 - \sqrt 3 - \left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right)}}\\
= \dfrac{{ - 2\sqrt 3 }}{{5 - 3}}\\
= - \sqrt 3 \\
d)\dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 3} \right) - \sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x + 3\sqrt x - x + 3\sqrt x }}{{x - 9}}\\
= \dfrac{{6\sqrt x }}{{x - 9}}\\
3)\\
b)Dkxd:x \ge - 2\\
\sqrt {16\left( {x + 2} \right)} = 8\\
\Rightarrow 4\sqrt {x + 2} = 8\\
\Rightarrow \sqrt {x + 2} = 2\\
\Rightarrow x + 2 = 4\\
\Rightarrow x = 2\left( {tmdk} \right)\\
\text{Vậy}\,x = 2\\
d)\sqrt {{x^2} + 6x + 9} - 3 = 0\\
\Rightarrow \sqrt {{{\left( {x + 3} \right)}^2}} = 3\\
\Rightarrow \left| {x + 3} \right| = 3\\
\Rightarrow \left[ \begin{array}{l}
x + 3 = 3\\
x + 3 = - 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = - 6
\end{array} \right.\\
\text{Vậy}\,x = 0;x = - 6
\end{array}$
