Đáp án:
$\begin{array}{l}
1)a)3\sqrt 8 + 4\sqrt {18} - 2\sqrt {50} \\
= 3.2\sqrt 2 + 4.3\sqrt 2 - 2.5\sqrt 2 \\
= 6\sqrt 2 + 12\sqrt 2 - 10\sqrt 2 \\
= 8\sqrt 2 \\
b)6\sqrt {20} - 2\sqrt {45} + 3\sqrt {80} \\
= 6.2\sqrt 5 - 2.3\sqrt 5 + 3.4\sqrt 5 \\
= 12\sqrt 5 - 6\sqrt 5 + 12\sqrt 5 \\
= 18\sqrt 5 \\
c)\sqrt {8 + 2\sqrt 7 } - \sqrt {11 + 4\sqrt 7 } \\
= \sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 7 + 2} \right)}^2}} \\
= \sqrt 7 + 1 - \left( {\sqrt 7 + 2} \right)\\
= - 1\\
d)\sqrt {22 - 6\sqrt {13} } - \sqrt {14 + 2\sqrt {13} } \\
= \sqrt {{{\left( {\sqrt {13} - 3} \right)}^2}} - \sqrt {{{\left( {\sqrt {13} + 1} \right)}^2}} \\
= \sqrt {13} - 3 - \sqrt {13} - 1\\
= - 4\\
2)a)Dkxd:x \ge - 4\\
\sqrt {9{x^2} - 6x + 1} = x + 4\\
\Leftrightarrow \sqrt {{{\left( {3x - 1} \right)}^2}} = x + 4\\
\Leftrightarrow \left| {3x - 1} \right| = x + 4\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 1 = x + 4\\
3x - 1 = - x - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\left( {tm} \right)\\
x = - \dfrac{3}{4}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{5}{2};x = - \dfrac{3}{4}\\
b)\sqrt {5 - 3x + 9{x^2}} - 2 = 3x\\
\Leftrightarrow \sqrt {9{x^2} - 3x + 5} = 3x - 2\left( {dk:x \ge \dfrac{2}{3}} \right)\\
\Leftrightarrow 9{x^2} - 3x + 5 = 9{x^2} - 12x + 4\\
\Leftrightarrow 9x = - 1\\
\Leftrightarrow x = - \dfrac{1}{9}\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
c)\sqrt {4{x^2} + 20x + 25} = \sqrt {16} \\
\Leftrightarrow 4{x^2} + 20x + 25 = 16\\
\Leftrightarrow {\left( {2x + 5} \right)^2} = {4^2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 5 = 4\\
2x + 5 = - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{1}{2}\\
x = - \dfrac{9}{2}
\end{array} \right.\\
Vậy\,x = - \dfrac{1}{2};x = - \dfrac{9}{2}
\end{array}$
