Bài 1:
a: Ta có: \(3\left(x-\dfrac{1}{2}\right)-3\left(x-\dfrac{1}{3}\right)=x\)
\(\Leftrightarrow x=3x-\dfrac{3}{2}-3x+1\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
b: Ta có: \(-\dfrac{4}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{3}{2}\left(2x-1\right)\)
\(\Leftrightarrow x\cdot\dfrac{-4}{3}+\dfrac{1}{3}-3x+\dfrac{3}{2}=0\)
\(\Leftrightarrow x\cdot\dfrac{-13}{3}=-\dfrac{11}{6}\)
hay \(x=\dfrac{11}{26}\)