a) |2x(x+3)|=x
\(< =>\left[{}\begin{matrix}2x=x\\x+3=x\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}2x-x=0\\x+3-x=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=0\\3=0\left(loai\right)\end{matrix}\right.\)
vậy x=0
b) |2-3x| = |\(\dfrac{1}{2}\)|
<=>\(\left[{}\begin{matrix}2-3x=\dfrac{-1}{2}\\2-3x=\dfrac{1}{2}\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}3x=\dfrac{5}{2}\\3x=\dfrac{3}{2}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy x=\(\dfrac{5}{6}\) và x=\(\dfrac{1}{2}\)