Đáp án:
$\begin{array}{l}
a)\left| {\dfrac{{3x - 6}}{{1 - 2x}}} \right| = x - 2\\
Dkxd:x - 2 \ge 0 \Leftrightarrow x \ge 2\\
TH1:\dfrac{{3x - 6}}{{1 - 2x}} = x - 2\\
\Leftrightarrow 3x - 6 = \left( {x - 2} \right)\left( {1 - 2x} \right)\\
\Leftrightarrow 3x - 6 = - 2{x^2} + 5x - 2\\
\Leftrightarrow 2{x^2} - 2x - 4 = 0\\
\Leftrightarrow {x^2} - x - 2 = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x = - 1\left( {ktm} \right)
\end{array} \right.\\
+ TH2:\\
\dfrac{{3x - 6}}{{1 - 2x}} = 2 - x\\
\Leftrightarrow 3x - 6 = \left( {1 - 2x} \right)\left( {2 - x} \right)\\
\Leftrightarrow 3x - 6 = 2{x^2} - 5x + 2\\
\Leftrightarrow 2{x^2} - 8x + 8 = 0\\
\Leftrightarrow {x^2} - 4x + 4 = 0\\
\Leftrightarrow x = 2\left( {tm} \right)\\
Vậy\,x = 2\\
b)Dkxd:\dfrac{{{x^2} - 6x + 8}}{{x + 3}} \ge 0\\
\left| { - 2x + 8} \right| = \dfrac{{{x^2} - 6x + 8}}{{x + 3}}\\
\Leftrightarrow 2\left| {x - 4} \right| = \dfrac{{\left( {x - 2} \right)\left( {x - 4} \right)}}{{x + 3}}\\
+ TH1:\\
2\left( {x - 4} \right) = \dfrac{{\left( {x - 2} \right)\left( {x - 4} \right)}}{{x + 3}}\\
\Leftrightarrow \left( {x - 4} \right).\left( {\dfrac{{x - 2}}{{x + 3}} - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\left( {tm} \right)\\
\dfrac{{x - 2}}{{x + 3}} = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x - 2 = 2x + 6 \Leftrightarrow x = - 8\left( {tm} \right)
\end{array} \right.\\
+ TH2:\\
- 2\left( {x - 4} \right) = \dfrac{{\left( {x - 2} \right)\left( {x - 4} \right)}}{{x + 3}}\\
\Leftrightarrow \left( {x - 4} \right).\left( {\dfrac{{x - 2}}{{x + 3}} + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
\dfrac{{x - 2}}{{x + 3}} = - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x - 2 = - 2x - 6 \Leftrightarrow x = - \dfrac{4}{3}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 4;x = - \dfrac{4}{3}\\
c)Dkxd:x\# 6;\# - 6\\
\dfrac{{\left| x \right| - 6}}{{{x^2} - 36}} = 2\\
\Leftrightarrow \left| x \right| - 6 = 2\left( {x - 6} \right)\left( {x + 6} \right)\\
+ Khi:x \ge 0\\
\Leftrightarrow x - 6 = 2\left( {x - 6} \right)\left( {x + 6} \right)\\
\Leftrightarrow \left( {x - 6} \right)\left( {2\left( {x + 6} \right) - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 6\left( {ktm} \right)\\
x + 6 = \dfrac{1}{2} \Leftrightarrow x = \dfrac{{ - 11}}{2}\left( {ktm} \right)
\end{array} \right.\\
+ Khi:x < 0\\
\Leftrightarrow - x - 6 = 2\left( {x - 6} \right)\left( {x + 6} \right)\\
\Leftrightarrow \left( {x + 6} \right)\left( {2\left( {x - 6} \right) + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 6\left( {ktm} \right)\\
x - 6 = - \dfrac{1}{2} \Leftrightarrow x = \dfrac{{11}}{2}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x \in \emptyset \\
d)Dkxd:5{x^2} - 7x + 2\# 0\\
\dfrac{{\left| {{x^2} - 4x + 3} \right|}}{{5{x^2} - 7x + 2}} = x - 3\\
\Leftrightarrow \dfrac{{\left| {\left( {x - 1} \right)\left( {x - 3} \right)} \right|}}{{\left( {x - 1} \right)\left( {5x - 2} \right)}} = x - 3\\
+ Khi:1 \le x \le 3\\
\Leftrightarrow \dfrac{{ - \left( {x - 1} \right)\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {5x - 2} \right)}} = x - 3\\
\Leftrightarrow \left( {x - 3} \right).\left( {\dfrac{1}{{5x - 2}} + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
5x - 2 = - 1 \Leftrightarrow x = \dfrac{1}{5}\left( {ktm} \right)
\end{array} \right.\\
+ Khi:\left[ \begin{array}{l}
x > 3\\
x < 1
\end{array} \right.\\
\Leftrightarrow \dfrac{{\left( {x - 1} \right)\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {5x - 2} \right)}} = x - 3\\
\Leftrightarrow \left( {x - 3} \right).\left( {\dfrac{1}{{5x - 2}} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\left( {ktm} \right)\\
5x - 2 = 1 \Leftrightarrow x = \dfrac{3}{5}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 3;x = \dfrac{3}{5}\\
f)\dfrac{{\left| {{x^2} + 5x + 4} \right|}}{{{x^2} + 3x + 2}} = x + 4\\
\Leftrightarrow \dfrac{{\left| {\left( {x + 1} \right)\left( {x + 4} \right)} \right|}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = x + 4\\
+ Khi: - 4 \le x \le - 1\\
\Leftrightarrow \dfrac{{ - \left( {x + 1} \right)\left( {x + 4} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = x + 4\\
\Leftrightarrow \left( {x + 4} \right)\left( {\dfrac{1}{{x + 2}} + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 4\left( {tm} \right)\\
\dfrac{1}{{x + 2}} = - 1 \Leftrightarrow x = - 3\left( {tm} \right)
\end{array} \right.\\
+ Khi:\left[ \begin{array}{l}
x > - 1\\
x < - 4
\end{array} \right.\\
\Leftrightarrow \dfrac{{\left( {x + 1} \right)\left( {x + 4} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = x + 4\\
\Leftrightarrow \left( {x + 4} \right).\left( {\dfrac{1}{{x + 2}} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 4\left( {ktm} \right)\\
x + 2 = 1 \Leftrightarrow x = - 1\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = - 4;x = - 3
\end{array}$
