Đáp án:
$\begin{array}{l}
a)\sqrt {4 - 2\sqrt 3 } .\sqrt {\sqrt 3 + 2} \\
= \sqrt {3 - 2.\sqrt 3 .1 + 1} .\sqrt {2\sqrt 3 + 4} .\dfrac{1}{{\sqrt 2 }}\\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} .\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} .\dfrac{1}{{\sqrt 2 }}\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {3 - 1} \right)\\
= \dfrac{1}{{\sqrt 2 }}.2\\
= \sqrt 2 \\
b)\sqrt {1 - 2x} - 2\sqrt {27x} + \sqrt {48x} \\
= \sqrt {1 - 2x} - 2.3\sqrt {3x} + 4\sqrt {3x} \\
= \sqrt {1 - 2x} - 2\sqrt {3x} \\
c)\sqrt {{x^2} - 4} - 2\sqrt {x + 2} = 0\left( {dkxd:x \ge 2} \right)\\
\Rightarrow \sqrt {\left( {x + 2} \right)\left( {x - 2} \right)} - 2\sqrt {x + 2} = 0\\
\Rightarrow \sqrt {x + 2} .\sqrt {x - 2} - 2\sqrt {x + 2} = 0\\
\Rightarrow \sqrt {x + 2} .\left( {\sqrt {x - 2} - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {x + 2} = 0\\
\sqrt {x - 2} = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x - 2 = 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 2\left( {ktm} \right)\\
x = 6\left( {tm} \right)
\end{array} \right.\\
Vay\,x = 6
\end{array}$
