a)
\(x+\dfrac{2}{3}=\dfrac{5}{6}\cdot\dfrac{3}{10}\\ x+\dfrac{2}{3}=\dfrac{1}{4}\\ x=\dfrac{1}{4}-\dfrac{2}{3}=\dfrac{-5}{12}\)
b)
\(\dfrac{7}{9}-\dfrac{5}{9}:x=\dfrac{1}{6}\\ \dfrac{5}{9}:x=\dfrac{7}{9}-\dfrac{1}{6}=\dfrac{11}{18}\\ x=\dfrac{5}{9}:\dfrac{11}{18}=\dfrac{10}{11}\)
c)
\(\left|x\right|+\dfrac{1}{5}=\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{2}\\x+\dfrac{1}{5}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}-\dfrac{1}{5}=\dfrac{13}{10}\\x=-\dfrac{3}{2}-\dfrac{1}{5}=-\dfrac{17}{10}\end{matrix}\right.\)
Vậy có 2 giá trị của \(x\) là: \(\dfrac{13}{10}\) và \(-\dfrac{17}{10}\)
