Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 5\\
\dfrac{{2x - 5}}{{x - 5}} = 3\\
\Rightarrow 2x - 5 = 3\left( {x - 5} \right)\\
\Rightarrow 2x - 5 = 3x - 15\\
\Rightarrow 3x - 2x = - 5 + 15\\
\Rightarrow x = 10\left( {tmdk} \right)\\
\text{Vậy}\,x = 10\\
b)\dfrac{{{x^2} - 6}}{x} = \dfrac{{2x + 2}}{3}\left( {dkxd:x \ne 0} \right)\\
\Rightarrow 3{x^2} - 18 = 2{x^2} + 2x\\
\Rightarrow {x^2} - 2x - 18 = 0\\
\Rightarrow {x^2} - 2x + 1 = 19\\
\Rightarrow {\left( {x - 1} \right)^2} = 19\\
\Rightarrow \left[ \begin{array}{l}
x = 1 + \sqrt {19} \left( {tm} \right)\\
x = 1 - \sqrt {19} \left( {tm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 1 \pm \sqrt {19} \\
c)\left| {x - 3} \right| = 2x + 1\left( {dk:x \ge - \dfrac{1}{2}} \right)\\
\Rightarrow \left[ \begin{array}{l}
x - 3 = 2x + 1\\
x - 3 = - 2x - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 4\left( {ktm} \right)\\
x = \dfrac{2}{3}\left( {tm} \right)
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{2}{3}
\end{array}$
