a, \(\frac{x}{2}-\frac{1}{x}=\frac{1}{12}\)
\(\Rightarrow\frac{x^2-2}{2x}=\frac{1}{12}\)
\(\Rightarrow12\left(x^2-2\right)=2x\)
\(\Rightarrow12x^2-24=2x\)
\(\Rightarrow12x^2-24-2x=0\)
\(\Rightarrow2\left(6x^2-x-12\right)=0\)
\(\Rightarrow6x^2-x-12=0\)
\(\Rightarrow6x^2+9x-8x-12=0\)
\(\Rightarrow3x\left(2x+3\right)-4\left(2x+3\right)=0\)
\(\Rightarrow\left(2x+3\right)\left(3x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=0\\3x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{4}{3}\end{matrix}\right.\)
Vậy x\(\in\left\{-\frac{3}{2};\frac{4}{3}\right\}\)
\(\frac{x}{2}-\frac{1}{x}=\frac{1}{12}\Rightarrow\frac{x^2-2}{2x}=\frac{1}{12}\Rightarrow12x^2-24=2x\\ \Rightarrow12x^2-2x-24=0\Rightarrow2\left(6x^2-x-12\right)=0\\ \Rightarrow6x^2-x-12=0\Rightarrow6x^2+8x-9x-12=0\\ \Rightarrow2x\left(3x+4\right)-3\left(3x+4\right)=0\Rightarrow\left(2x-3\right)\left(3x+4\right)=0\)
*\(2x-3=0\Rightarrow x=\frac{3}{2}\)
*\(3x+4=0\Rightarrow x=\frac{-4}{3}\)