Điều kiện: \(x\ne0\)
\(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\\ \Leftrightarrow6x^2-12-x=0\\ \Leftrightarrow6x^2-9x+8x-12=0\\ \Leftrightarrow3x\left(2x-3\right)+4\left(2x-3\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{3}{2}\end{matrix}\right.\left(tm\right)}\)