$a)$ $|5x - 4| = |x + 2|$
$⇒$ \(\left[ \begin{array}{l}5x-4=x+2\\5x-4=-x-2\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}4x=6\\6x=2\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy $x$ $∈$ `{3/2;1/3}`.
$b$) `|2x - 3| - |3x + 2| = 0`
`⇔ |2x-3| = |3x+2|`
$⇒$ \(\left[ \begin{array}{l}2x-3=3x+2\\2x-3=-3x-2\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}-x=5\\5x=1\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=-5\\x=\dfrac{1}{5}\end{array} \right.\)
Vậy $x$ $∈$ `{-5;1/5}`.
