\(2.\left|2x-3\right|=\frac{1}{2}\)
=> \(\left|2x-3\right|=\frac{1}{4}\)
=> 2x - 3 = + \(\frac{1}{4}\)
+ TH1: 2x - 3 = \(\frac{1}{4}\)
=> 2x = \(\frac{13}{4}\)
=> x = \(\frac{13}{8}\)
+ TH2: 2x - 3 = \(-\frac{1}{4}\)
=> 2x = \(\frac{11}{4}\)
=> x = \(\frac{11}{8}\)
KL: \(x\in\left\{\frac{13}{8};\frac{11}{8}\right\}\)