\(e,\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x\right|=\frac{20}{12}+\frac{9}{12}\)
\(\left|x\right|=\frac{11}{12}\)
⇒\(x=\frac{11}{12}\)hoặc\(x=\frac{-11}{12}\)
\(f,\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)
\(\left|2x-\frac{1}{3}\right|=1-\frac{5}{6}\)
\(\left|2x-\frac{1}{3}\right|=\frac{1}{6}\)
\(\Rightarrow2x-\frac{1}{3}=\frac{1}{6}\)hoặc \(2x-\frac{1}{3}=\frac{-1}{6}\)
Ta có 2 trường hợp.
\(TH1.2x-\frac{1}{3}=\frac{1}{6}\) \(TH2.2x-\frac{1}{3}=\frac{-1}{6}\)
\(2x=\frac{1}{6}+\frac{1}{3}\) \(2x=\frac{-1}{6}+\frac{1}{3}\)
\(2x=\frac{1}{6}+\frac{2}{6}\) \(2x=\frac{-1}{6}+\frac{2}{6}\)
\(x=\frac{1}{2}:2\) \(x=\frac{-1}{6}:2\)
\(x=\frac{1}{4}\) \(x=\frac{1}{14}\)