Đáp án:
$\begin{array}{l}
1)\left| {{x^2} - 2x} \right| = \left| {x - 2} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2x = x - 2\\
{x^2} - 2x = 2 - x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 3x + 2 = 0\\
{x^2} - x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {x - 1} \right)\left( {x - 2} \right) = 0\\
\left( {x - 2} \right)\left( {x + 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = - 1
\end{array} \right.\\
Vậy\,x = 1;x = 2;x = - 1\\
2)\left| {2x - 1} \right| + x = 1 + {x^2}\\
\Leftrightarrow \left| {2x - 1} \right| = {x^2} - x + 1\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - x + 1 = 2x - 1\\
{x^2} - x + 1 = - 2x + 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 3x + 2 = 0\\
{x^2} + x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {x - 1} \right)\left( {x - 2} \right) = 0\\
x\left( {x + 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = 0\\
x = - 1
\end{array} \right.\\
Vậy\,x \in \left\{ { - 1;0;1;2} \right\}\\
3)\\
3\left| x \right| = 2x + 1\left( {dk:x \ge - \dfrac{1}{2}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 2x + 1\\
- 3x = 2x + 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
- 5x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{5}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 1
\end{array}$
