Tìm x: 

- Hướng dẫn giải

$$\begin{gathered} \frac{x}{2}-\frac{1}{x}=\frac{1}{12} \\ \begin{array}{l}\frac{x^2-2}{2x}=\frac{1}{12}\\ 12\left(x^2-2\right)=2x\\ 6\left(x^2-2\right)=x\\ 6x^2-x-12=0\\ \begin{array}{l}6x^2+8x-9x-12=0\\ 2x\left(3x+4\right)-3\left(3x+4\right)=0\\ \left(3x+4\right)\left(2x-3\right)=0\end{array}\end{array} \end{gathered}$$

TH1: 3x + 4 = 0

$\begin{array}{l}3x=-4\\ x=-\frac{4}{3}\end{array}$

TH2: 2x - 3 = 0

$$\begin{gathered} \begin{array}{l}2x=3\\ x=\frac{3}{2}\end{array} \\ Vậy x\in \left\{\frac{-4}{3};\frac{3}{2}\right\} \end{gathered}$$