Ta có: f(x)=a⇔|3x-2|=a⇔\(\left[ \begin{array}{l}3x-2=a\\3x-2=-a\end{array} \right.\) ⇔\(\left[ \begin{array}{l}3x=2+a\\3x=2-a\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=\frac{2+a}{3}\\x=\frac{2-a}{3}\end{array} \right.\)
Khi a=0: x=$\frac{2}{3}$
Khi a=1: \(\left[ \begin{array}{l}x=1\\x=\frac{1}{3}\end{array} \right.\)
Khi a=\frac{1}{4}: \(\left[ \begin{array}{l}x=\frac{3}{4}\\x=\frac{7}{12}\end{array} \right.\)
khi a=2019: \(\left[ \begin{array}{l}x=\frac{2021}{3}\\x=\frac{-2017}{3}\end{array} \right.\)
