\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4\left(x-2\right)\right|\\ \Rightarrow\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-8\right|\\ \Rightarrow\left[{}\begin{matrix}\frac{3}{2}x+\frac{1}{2}=4x-8\\\frac{3}{2}x+\frac{1}{2}=-4x+8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x-\frac{3}{2}x=8+\frac{1}{2}\\\frac{3}{2}x+4x=8-\frac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{5}{2}x=\frac{17}{2}\\\frac{11}{2}x=\frac{15}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{17}{5}\\x=\frac{15}{11}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{17}{5};\frac{15}{11}\right\}\)