Đáp án:
$\begin{array}{l}
a)3\left( {x - 1} \right) - 2x\left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {3 - 2x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
3 - 2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{3}{2}
\end{array} \right.\\
\text{Vậy}\,x = 1;x = \dfrac{3}{2}\\
b)5\left( {x + 2} \right) - 4x\left( {2 - x} \right) = 0\\
\Leftrightarrow 5x + 10 - 8x + 4{x^2} = 0\\
\Leftrightarrow 4{x^2} - 3x + 10 = 0\left( {vn} \right)\\
\text{Vậy pt vô nghiệm}\\
c)x\left( {2x - 4} \right) = \left( {3x - 1} \right)\left( {2x - 4} \right)\\
\Leftrightarrow \left( {2x - 4} \right)\left( {x - 3x + 1} \right) = 0\\
\Leftrightarrow 2.\left( {x - 2} \right).\left( {1 - 2x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
1 - 2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \dfrac{1}{2}
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{1}{2};x = 2\\
d){x^2} - x + 1 = {x^2} - x - 1\\
\Leftrightarrow 1 = - 1\left( {ktm} \right)\\
\text{Vậy pt vô nghiệm}\\
e){x^2} - 7x + 12 = 0\\
\Leftrightarrow {x^2} - 4x - 3x + 12 = 0\\
\Leftrightarrow \left( {x - 4} \right)\left( {x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 3
\end{array} \right.\\
\text{Vậy}\,x = 3;x = 4\\
f)Dkxd:x\# \dfrac{3}{2}\\
\dfrac{{\left( {x + 2} \right)}}{{2x - 3}} = \dfrac{{1 - x}}{{3 - 2x}} + 1\\
\Leftrightarrow \dfrac{{x + 2}}{{2x - 3}} - \dfrac{{x - 1}}{{2x - 3}} = 1\\
\Leftrightarrow \dfrac{{x + 2 - x + 1}}{{2x - 3}} = 1\\
\Leftrightarrow \dfrac{3}{{2x - 3}} = 1\\
\Leftrightarrow 2x - 3 = 3\\
\Leftrightarrow x = 3\left( {tmdk} \right)\\
\text{Vậy}\,x = 3\\
g)Dkxd:x\# 2;x\# - 2\\
\dfrac{{x + 2}}{{x - 2}} - \dfrac{{x - 2}}{{x + 2}} = \dfrac{{ - 1}}{{4 - {x^2}}}\\
\Leftrightarrow \dfrac{{{{\left( {x + 2} \right)}^2} - {{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\Leftrightarrow {x^2} + 4x + 4 - {x^2} + 4x - 4 = 1\\
\Leftrightarrow 8x = 1\\
\Leftrightarrow x = \dfrac{1}{8}\left( {tmdk} \right)\\
\text{Vậy}\,x = \dfrac{1}{8}
\end{array}$
