Ta có: \(\frac{1}{2}\left(3x-\frac{1}{2}\right)-\frac{1}{4}x=7\)
\(\Leftrightarrow\frac{3}{2}x-\frac{1}{4}-\frac{1}{4}x=7\)
\(\Leftrightarrow\frac{5}{4}x-\frac{1}{4}=7\)
\(\Leftrightarrow\frac{5}{4}x=7+\frac{1}{4}=\frac{29}{4}\)
hay \(x=\frac{29}{4}:\frac{5}{4}=\frac{29}{4}\cdot\frac{4}{5}=\frac{29}{5}\)
Vậy: \(x=\frac{29}{5}\)