Đáp án:
$\begin{array}{l}
c)2\left| {x - 2} \right| + 3x = 0\\
\Leftrightarrow 2\left| {x - 2} \right| = - 3x\left( {dk:x \le 0} \right)\\
\Leftrightarrow 2.\left( {2 - x} \right) = - 3x\left( {khi:x \le 0 \Leftrightarrow x - 2 < 0} \right)\\
\Leftrightarrow 4 - 2x = - 3x\\
\Leftrightarrow - 2x + 3x = - 4\\
\Leftrightarrow x = - 4\left( {tm} \right)\\
Vậy\,x = - 4\\
e)5 - 2\left| {3 - 2x} \right| = x + 2\\
\Leftrightarrow 2\left| {3 - 2x} \right| = 5 - x - 2\\
\Leftrightarrow 2\left| {3 - 2x} \right| = 3 - x\left( {dk:x \le 3} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2\left( {3 - 2x} \right) = 3 - x\\
2\left( {2x - 3} \right) = 3 - x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
6 - 4x = 3 - x \Leftrightarrow x = 1\left( {tm} \right)\\
4x - 6 = 3 - x \Leftrightarrow x = \dfrac{9}{5}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 1;x = \dfrac{9}{5}
\end{array}$
