Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge \dfrac{1}{2}\\
3\sqrt {2x - 1} - 4\sqrt {2x - 1} = - 2\\
\Leftrightarrow - \sqrt {2x - 1} = - 2\\
\Leftrightarrow \sqrt {2x - 1} = 2\\
\Leftrightarrow 2x - 1 = 4\\
\Leftrightarrow 2x = 5\\
\Leftrightarrow x = \dfrac{5}{2}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{5}{2}\\
b)Dkxd:x \le \dfrac{7}{3}\\
5\sqrt {7 - 3x} - 12 = 2\sqrt {7 - 3x} \\
\Leftrightarrow 3\sqrt {7 - 3x} = 12\\
\Leftrightarrow \sqrt {7 - 3x} = 4\\
\Leftrightarrow 7 - 3x = 16\\
\Leftrightarrow 3x = - 11\\
\Leftrightarrow x = - \dfrac{{11}}{3}\left( {tmdk} \right)\\
Vậy\,x = - \dfrac{{11}}{3}\\
c)Dkxd:x \ge \dfrac{{ - 4}}{5}\\
16 - \sqrt {4 + 5x} = 3\sqrt {4 + 5x} \\
\Leftrightarrow 4\sqrt {4 + 5x} = 16\\
\Leftrightarrow \sqrt {4 + 5x} = 4\\
\Leftrightarrow 4 + 5x = 16\\
\Leftrightarrow 5x = 12\\
\Leftrightarrow x = \dfrac{{12}}{5}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{12}}{5}\\
d)\sqrt {{{\left( {2x - 3} \right)}^2}} = 18\\
\Leftrightarrow \left| {2x - 3} \right| = 18\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 18\\
2x - 3 = - 18
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{21}}{2}\\
x = - \dfrac{{15}}{2}
\end{array} \right.\\
Vậy\,x = \dfrac{{21}}{2};x = - \dfrac{{15}}{2}
\end{array}$
