Đáp án:
$\begin{array}{l}
a)2x - 2 = 4\left( {x - 1} \right)\\
\Rightarrow 2\left( {x - 1} \right) = 4\left( {x - 1} \right)\\
\Rightarrow x = 1\\
b)3x\left( {x - 2} \right) - 2x + 4 = 0\\
\Rightarrow 3x\left( {x - 2} \right) - 2\left( {x - 2} \right) = 0\\
\Rightarrow \left( {x - 2} \right)\left( {3x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x - 2 = 0\\
3x - 2 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = \frac{2}{3}
\end{array} \right.\\
c)\frac{{x + 2}}{{x - 2}} - \frac{1}{x} = \frac{2}{{{x^2} - 2x}}\left( {dk:x \ne 0;x \ne 2} \right)\\
\Rightarrow \frac{{x\left( {x + 2} \right) - x + 2}}{{x\left( {x - 2} \right)}} = \frac{2}{{x\left( {x - 2} \right)}}\\
\Rightarrow {x^2} + 2x - x + 2 = 2\\
\Rightarrow {x^2} + x = 0\\
\Rightarrow x\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\left( {ktm} \right)\\
x = - 1\left( {tm} \right)
\end{array} \right.\\
Vay\,x = - 1\\
d)\left| {x - 3} \right| = 3x + 1\left( {x \ge - \frac{1}{3}} \right)\\
\Rightarrow \left[ \begin{array}{l}
x - 3 = 3x + 1\\
x - 3 = - 3x - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 2\left( {ktm} \right)\\
x = \frac{1}{2}\left( {tm} \right)
\end{array} \right.\\
Vay\,x = \frac{1}{2}\\
e)\frac{{2x + 6}}{4} - 2 \le \frac{x}{3}\\
\Rightarrow \frac{{x + 3}}{2} - 2 \le \frac{x}{3}\\
\Rightarrow \frac{{x + 3 - 4}}{2} \le \frac{x}{3}\\
\Rightarrow \frac{{x - 1}}{2} - \frac{x}{3} \le 0\\
\Rightarrow \frac{{3x - 3 - 2x}}{6} \le 0\\
\Rightarrow x - 3 \le 0\\
\Rightarrow x \le 3
\end{array}$
