Bài 2:
a: Ta có: \(x-\dfrac{4}{3}=\dfrac{-2}{5}\)
\(\Leftrightarrow x=\dfrac{-2}{5}+\dfrac{4}{3}=\dfrac{14}{15}\)
b: Ta có: \(x\cdot\dfrac{-5}{2}-\dfrac{2}{5}=\dfrac{-2}{3}\)
\(\Leftrightarrow x\cdot\dfrac{-5}{2}=\dfrac{-4}{15}\)
hay \(x=\dfrac{8}{75}\)


