a) \(\frac{3}{4}x-\frac{1}{4}=2\left(x-3\right)+\frac{1}{4}x\)
\(\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\frac{3}{4}x-2x-\frac{1}{4}x=\frac{1}{4}-6\)
\(x\left(\frac{3}{4}-2-\frac{1}{4}\right)=-\frac{23}{4}\)
\(-\frac{3}{2}x=-\frac{23}{4}\)
\(x=-\frac{23}{4}\div\left(-\frac{3}{2}\right)\)
\(x=\frac{23}{6}\)
b) \(30\%-x+\frac{5}{6}=\frac{1}{3}\)
\(\frac{3}{10}-x+\frac{5}{6}=\frac{1}{3}\)
\(\frac{3}{10}-x=\frac{1}{3}-\frac{5}{6}\)
\(\frac{3}{10}-x=-\frac{1}{2}\)
\(x=\frac{3}{10}-\left(-\frac{1}{2}\right)\)
\(x=\frac{4}{5}\)
c) \(\left(2x-1\right).\left(3x+12\right)=0\)
\(\hept{\begin{cases}2x-1=0\\3x+12=0\end{cases}\Rightarrow\hept{\begin{cases}2x=1\\3x=-12\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\x=-4\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{2};-4\right\}\)
b) \(30\%x-x+\frac{5}{6}=\frac{1}{3}\)
\(\frac{3}{10}x-x+\frac{5}{6}=\frac{1}{3}\)
\(\frac{3}{10}x-1=\frac{1}{3}-\frac{5}{6}\)
\(\frac{3}{10}x-x=-\frac{1}{2}\)
\(x\left(\frac{3}{10}-1\right)=-\frac{1}{2}\)
\(-\frac{7}{10}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\left(-\frac{7}{10}\right)\)
\(x=\frac{5}{7}\)
d) \(|2x-\frac{1}{2}|=\frac{3}{4}\)
\(\hept{\begin{cases}2x-\frac{1}{2}=\frac{3}{4}\\2x-\frac{1}{2}=-\frac{3}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{5}{4}\\2x=-\frac{1}{4}\end{cases}\Rightarrow}\hept{\begin{cases}2x=\frac{5}{8}\\2x=-\frac{1}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{5}{8};-\frac{1}{8}\right\}\)
