Ta có: \(\dfrac{1}{2}\left(2x+3\right)+\dfrac{1}{3}x=\dfrac{1}{12}\)
\(\Leftrightarrow x+\dfrac{3}{2}+\dfrac{1}{3}x=\dfrac{1}{12}\)
\(\Leftrightarrow\dfrac{4}{3}x=\dfrac{1}{12}-\dfrac{3}{2}=\dfrac{1}{12}-\dfrac{18}{12}=\dfrac{-17}{12}\)
\(\Leftrightarrow x=\dfrac{-17}{12}:\dfrac{4}{3}=\dfrac{-17}{12}\cdot\dfrac{3}{4}=\dfrac{-51}{48}=\dfrac{-17}{16}\)
Vậy: \(x=-\dfrac{17}{16}\)