Đáp án:

$\begin{array}{l}
a)\sqrt {{{\left( {3x - 2} \right)}^2}}  = 4\\
 \Leftrightarrow \left| {3x - 2} \right| = 4\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - 2 = 4\\
3x - 2 =  - 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{2}{3}
\end{array} \right.\\
Vậy\,x = 2;x =  - \dfrac{2}{3}\\
b)Dkxd:x \ge 0\\
\dfrac{5}{3}\sqrt {15x}  - \sqrt {15x}  - 2 = \dfrac{1}{3}\sqrt {15x} \\
 \Leftrightarrow \dfrac{5}{3}\sqrt {15x}  - \sqrt {15x}  - \dfrac{1}{3}\sqrt {15x}  = 2\\
 \Leftrightarrow \dfrac{1}{3}\sqrt {15x}  = 2\\
 \Leftrightarrow \sqrt {15x}  = 6\\
 \Leftrightarrow 15x = 36\\
 \Leftrightarrow x = \dfrac{{36}}{{15}} = \dfrac{{12}}{5}\\
Vậy\,x = \dfrac{{12}}{5}\\
c)Dkxd:x \ge 3\\
\sqrt {4x - 12}  + \sqrt {9x - 27}  - 4\sqrt {x - 3}  + 3 - x = 0\\
 \Leftrightarrow 2\sqrt {x - 3}  + 3\sqrt {x - 3}  - 4\sqrt {x - 3}  = x - 3\\
 \Leftrightarrow \sqrt {x - 3}  = x - 3\\
 \Leftrightarrow \sqrt {x - 3} \left( {\sqrt {x - 3}  - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 3}  = 0\\
\sqrt {x - 3}  = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = 4\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 3;x = 4\\
d)Dkxd:x \ge 2\\
\sqrt {25x + 75}  + 3\sqrt {x - 2}  = 2 + 4\sqrt {x + 3}  + \sqrt {9x - 18} \\
 \Leftrightarrow 5\sqrt {x + 3}  + 3\sqrt {x - 2}  = 2 + 2\sqrt {x + 3}  + 3\sqrt {x - 2} \\
 \Leftrightarrow 3\sqrt {x + 3}  = 2\\
 \Leftrightarrow x + 3 = \dfrac{4}{9}\\
 \Leftrightarrow x = \dfrac{4}{9} - 3 = \dfrac{{ - 23}}{9}\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
e)Dkxd:x \ge \dfrac{1}{3}\\
1 + \sqrt {3x + 1}  = 3x\\
 \Leftrightarrow \sqrt {3x + 1}  = 3x - 1\\
 \Leftrightarrow 3x + 1 = 9{x^2} - 6x + 1\\
 \Leftrightarrow 9{x^2} - 9x = 0\\
 \Leftrightarrow 9x\left( {x - 1} \right) = 0\\
 \Leftrightarrow x = 1\left( {do:x \ge \dfrac{1}{3}} \right)\\
Vậy\,x = 1\\
9)\\
A = \dfrac{{3x + \sqrt {9x}  - 3}}{{x + \sqrt x  - 2}} - \dfrac{{\sqrt x  + 1}}{{\sqrt x  + 2}} + \dfrac{{\sqrt x  + 2}}{{1 - \sqrt x }}\\
 = \dfrac{{3x + 3\sqrt x  - 3 - \left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right) - {{\left( {\sqrt x  + 2} \right)}^2}}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{3x + 3\sqrt x  - 3 - x + 1 - x - 4\sqrt x  - 4}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{x - \sqrt x  - 6}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x  - 3}}{{\sqrt x  - 1}}
\end{array}$