Đáp án:
$\begin{array}{l}
1a)\sqrt {75} + \sqrt {27} - \sqrt {48} + \sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} \\
= 5\sqrt 3 + 3\sqrt 3 - 4\sqrt 3 + \sqrt 3 - 1\\
= 5\sqrt 3 - 1\\
b)\dfrac{1}{{2 - \sqrt 3 }} - \dfrac{{\sqrt 6 - \sqrt 2 }}{{\sqrt 3 - 1}} - \sqrt {5 - 2\sqrt 6 } \\
= \dfrac{{2 + \sqrt 3 }}{{{2^2} - 3}} - \dfrac{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 - 1}} - \sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \\
= 2 + \sqrt 3 - \sqrt 2 - \left( {\sqrt 3 - \sqrt 2 } \right)\\
= 2\\
2)a)\sqrt {{x^2} - 2x + 1} = 3\\
\Leftrightarrow {x^2} - 2x + 1 = 9\\
\Leftrightarrow {x^2} - 2x - 8 = 0\\
\Leftrightarrow \left( {x - 4} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow x = 4;x = - 2\\
Vậy\,x = 4;x = - 2\\
b)Dkxd:x \ge - 3\\
\sqrt {4x + 12} - 3\sqrt {3 + x} + 7\sqrt {9x + 27} = 20\\
\Leftrightarrow 2\sqrt {x + 3} - 3\sqrt {x + 3} + 7.3\sqrt {x + 3} = 20\\
\Leftrightarrow 20\sqrt {x + 3} = 20\\
\Leftrightarrow \sqrt {x + 3} = 1\\
\Leftrightarrow x + 3 = 1\\
\Leftrightarrow x = - 2\left( {tmdk} \right)\\
Vậy\,x = - 2
\end{array}$
