Câu 1:
a) Ta có: \(\left(x-\frac{5}{8}\right)\cdot\frac{5}{18}=\frac{15}{36}\)
\(\Leftrightarrow x-\frac{5}{8}=\frac{15}{36}:\frac{5}{8}=\frac{15}{36}\cdot\frac{8}{5}=\frac{120}{180}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}+\frac{5}{8}=\frac{16}{24}+\frac{15}{24}=\frac{41}{24}\)
Vậy: \(x=\frac{41}{24}\)
b) Ta có: \(\left|x-\frac{1}{3}\right|=\frac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{5}{6}\\x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{6}+\frac{1}{3}=\frac{5}{6}+\frac{2}{6}=\frac{7}{6}\\x=\frac{-5}{6}+\frac{1}{3}=\frac{-5}{6}+\frac{2}{6}=\frac{-3}{6}=\frac{-1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{7}{6};\frac{-1}{2}\right\}\)
