a) \(2\left|\frac{3}{2}x-\frac{1}{4}\right|=\left|-\frac{5}{4}\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2\left|\frac{3}{2}x-\frac{1}{4}\right|=\frac{5}{4}\\2\left|\frac{3}{2}x-\frac{1}{4}\right|=-\frac{5}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left|\frac{3}{2}x-\frac{1}{4}\right|=\frac{5}{4}:2\\\left|\frac{3}{2}x-\frac{1}{4}\right|=-\frac{5}{4}:2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left|\frac{3}{2}x-\frac{1}{4}\right|=\frac{5}{8}\\\left|\frac{3}{2}x-\frac{1}{4}\right|=-\frac{5}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x-\frac{1}{4}=\frac{5}{8}\\\frac{3}{2}x-\frac{1}{4}=-\frac{5}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x=\frac{5}{8}+\frac{1}{4}\\\frac{3}{2}x=-\frac{5}{8}+\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x=\frac{7}{8}\\\frac{3}{2}x=-\frac{3}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}:\frac{3}{2}\\x=-\frac{3}{8}:\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}.\frac{2}{3}\\x=-\frac{3}{8}.\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=-\frac{1}{4}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{12};-\frac{1}{4}\right\}\)
