$\begin{array}{l} 81)\\ \left| {x - 2} \right| = \left| {3x - 5} \right|\\  \Leftrightarrow \left[ \begin{array}{l} x - 2 = 3x - 5\\ x - 2 = 5 - 3x \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} 2x = 3\\ 4x = 7 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{3}{2}\\ x = \dfrac{7}{4} \end{array} \right.\\  \Rightarrow S = \left\{ {\dfrac{3}{2};\dfrac{7}{4}} \right\} \end{array}$

$\begin{array}{l} 82)\left| {x + 2} \right| = 2\left| {x - 2} \right|\\  \Leftrightarrow \left[ \begin{array}{l} x + 2 = 2\left( {x - 2} \right)\\ x + 2 = 2\left( {2 - x} \right) \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x + 2 = 2x - 4\\ x + 2 = 4 - 2x \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = 6\\ x = \dfrac{2}{3} \end{array} \right. \Rightarrow S = 6 + \dfrac{2}{3} = \dfrac{{20}}{3} \end{array}$

$\begin{array}{l} 83)\left| {2x + 1} \right| = \left| {{x^2} - 3x - 4} \right|\\  \Leftrightarrow \left[ \begin{array}{l} 2x + 1 = {x^2} - 3x - 4\\ 2x + 1 =  - \left( {{x^2} - 3x - 4} \right) \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} {x^2} - 5x - 5 = 0\\ 2x + 1 =  - {x^2} + 3x + 4 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} {x^2} - 5x - 5 = 0\\  - {x^2} + x + 3 = 0 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{5 \pm \sqrt 5 }}{2}\\ x = \dfrac{{1 \pm \sqrt {13} }}{2} \end{array} \right. \to 4 \to D \end{array}$

$\begin{array}{l} 84) \left| {2x - 4} \right| + \left| {x - 1} \right| = 0\\  \Leftrightarrow \left\{ \begin{array}{l} 2x - 4 = 0\\ x - 1 = 0 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ x = 1 \end{array} \right.(L) \to A \end{array}$

$\begin{array}{l}
85)\left| {2x - 5} \right| + \left| {2{x^2} - 7x + 5} \right| = 0\\
 \Leftrightarrow \left| {2x - 5} \right| + \left| {\left( {2x - 5} \right)\left( {x - 1} \right)} \right| = 0\\
 \Leftrightarrow \left| {2x - 5} \right|\left( {\underbrace {1 + \left| {x - 1} \right|}_{ \ge 1}} \right) = 0\\
 \Leftrightarrow 2x - 5 = 0\\
 \Leftrightarrow x = \dfrac{5}{2} \to B
\end{array}$

$\begin{array}{l} 86)\\ {\left( {x + 1} \right)^2} - 3\left| {x + 1} \right| + 2 = 0\\  \Leftrightarrow {\left( {\left| {x + 1} \right|} \right)^2} - 3\left| {x + 1} \right| + 2 = 0\\  \Leftrightarrow \left( {\left| {x + 1} \right| - 2} \right)\left( {\left| {x + 1} \right| - 1} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} \left| {x + 1} \right| = 2\\ \left| {x + 1} \right| = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x + 1 = 2\\ x + 1 = 1\\ x + 1 =  - 1\\ x + 1 =  - 2 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = 0\\ x =  - 2\\ x =  - 3 \end{array} \right. \to D \end{array}$