Giải thích các bước giải:
$\begin{array}{l}
a)2\left| {2x - 3} \right| = 2x + 1\\
\Leftrightarrow \left\{ \begin{array}{l}
2x + 1 \ge 0\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
2\left( {2x - 3} \right) = 2x + 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x < \dfrac{3}{2}\\
2\left( { - 2x + 3} \right) = 2x + 1
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{{ - 1}}{2}\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x = \dfrac{7}{2}
\end{array} \right.\left( {tm} \right)\\
\left\{ \begin{array}{l}
x < \dfrac{3}{2}\\
x = \dfrac{5}{6}
\end{array} \right.\left( {tm} \right)
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x = \dfrac{5}{6}
\end{array} \right.
\end{array}$
Vậy tập nghiệm của phương trình là $S = \left\{ {\dfrac{5}{6};\dfrac{7}{2}} \right\}$
$\begin{array}{l}
b)3\left( {x - 1} \right)\sqrt {x + 1} - 2 = 2{x^2} - 4x\left( {DK:x \ge - 1} \right)\\
\Leftrightarrow 3\left( {x - 1} \right)\sqrt {x + 1} - 2\left( {{x^2} - 2x + 1} \right) = 0\\
\Leftrightarrow 3\left( {x - 1} \right)\sqrt {x + 1} - 2{\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {3\sqrt {x + 1} - 2\left( {x - 1} \right)} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
3\sqrt {x + 1} - 2\left( {x - 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
3\sqrt {x + 1} = 2\left( {x - 1} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
x - 1 \ge 0\\
9\left( {x - 1} \right) = 4{\left( {x - 1} \right)^2}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
x \ge 1\\
4{x^2} - 17x + 13 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
x \ge 1\\
\left( {x - 1} \right)\left( {4x - 13} \right) = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x = 1\\
x = \dfrac{{13}}{4}
\end{array} \right.\left( {tm} \right)
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{13}}{4}
\end{array} \right.\left( {tm} \right)
\end{array}$
Vậy tập nghiệm của phương trình là $S = \left\{ {1;\dfrac{{13}}{4}} \right\}$
