Đáp án:
$\begin{array}{l}
1)\\
a)\left( {\sqrt {18} + \sqrt {20} - \sqrt 8 } \right).\sqrt 2 - 2\sqrt {10} \\
= \sqrt {36} + \sqrt {40} - \sqrt {16} - 2\sqrt {10} \\
= 6 + 2\sqrt {10} - 4 - 2\sqrt {10} \\
= 2\\
b)\sqrt {{{\left( {\sqrt 3 - 3} \right)}^2}} - \sqrt {12 + 6\sqrt 3 } \\
= 3 - \sqrt 3 - \sqrt {9 + 2.3.\sqrt 3 + 3} \\
= 3 - \sqrt 3 - \sqrt {{{\left( {3 + \sqrt 3 } \right)}^2}} \\
= 3 - \sqrt 3 - 3 - \sqrt 3 \\
= - 2\sqrt 3 \\
c)\dfrac{3}{{\sqrt 5 - \sqrt 2 }} + \dfrac{2}{{2 + \sqrt 2 }} + \dfrac{{\sqrt 5 - 5}}{{\sqrt 5 - 1}}\\
= \dfrac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{{5 - 2}} + \dfrac{{\sqrt 2 \left( {\sqrt 2 - 1} \right)}}{{2 - 1}} + \dfrac{{\sqrt 5 \left( {1 - \sqrt 5 } \right)}}{{\sqrt 5 - 1}}\\
= \sqrt 5 + \sqrt 2 + 2 - \sqrt 2 - \sqrt 5 \\
= 2\\
2)a)Dkxd:x \ge - 3\\
\sqrt {x + 3} + 2\sqrt {4x + 12} - \dfrac{1}{3}\sqrt {9x + 27} = 8\\
\sqrt {x + 3} + 4\sqrt {x + 3} - \sqrt {x + 3} = 8\\
4\sqrt {x + 3} = 8\\
\sqrt {x + 3} = 2\\
x + 3 = 4\\
\Rightarrow x = 1\left( {tmdk} \right)\\
b)\sqrt {4{x^2} - 4x + 1} = \sqrt {{x^2} + 10x + 25} \\
\Rightarrow 4{x^2} - 4x + 1 = {x^2} + 10x + 25\\
\Rightarrow 3{x^2} - 14x - 24 = 0\\
\Rightarrow \left( {x - 6} \right)\left( {3x + 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 6\\
x = - \dfrac{4}{3}
\end{array} \right.
\end{array}$
