Câu 2:
a: Ta có: \(x-\dfrac{1}{3}=\dfrac{6}{19}+\dfrac{-3}{9}\)
\(\Leftrightarrow x-\dfrac{1}{3}=\dfrac{6}{19}-\dfrac{1}{3}\)
hay \(x=\dfrac{6}{19}\)
b: Ta có: \(\dfrac{1}{x}\cdot\dfrac{-2}{7}=\dfrac{3}{8}\)
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{3}{8}:\dfrac{-2}{7}=\dfrac{21}{-16}=\dfrac{-21}{16}\)
hay \(x=-\dfrac{16}{21}\)