Bài 1:
a) Ta có: \(\left|2x-\frac{1}{2}\right|=\frac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{1}{2}=\frac{5}{4}\\2x-\frac{1}{2}=\frac{-5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{5}{4}+\frac{1}{2}=\frac{7}{4}\\2x=\frac{-5}{4}+\frac{1}{2}=\frac{-3}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{4}:2=\frac{7}{8}\\x=\frac{-3}{4}:2=\frac{-3}{8}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{7}{8};\frac{-3}{8}\right\}\)
b) Ta có: \( \left|3-2x\right|=\frac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}3-2x=\frac{1}{3}\\3-2x=\frac{-1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=\frac{1}{3}-3=\frac{-8}{3}\\-2x=\frac{-1}{3}-3=\frac{-10}{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-8}{3}:\left(-2\right)=\frac{-8}{3}\cdot\frac{1}{-2}=\frac{4}{3}\\x=\frac{-10}{3}:\left(-2\right)=-\frac{10}{3}\cdot\frac{1}{-2}=\frac{5}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{4}{3};\frac{5}{3}\right\}\)
