Tìm x: 

- Hướng dẫn giải

$\left(2x-\frac{1}{2}\right).\left(3x-\frac{1}{3}\right)<0$

TH1: $\left\{\begin{array}{l}2x-\frac{1}{2}>0\\ 3x-\frac{1}{3}<0\end{array}\right.\iff \left\{\begin{array}{l}2x>\frac{1}{2}\\ 3x<\frac{1}{3}\end{array}\right.\iff \left\{\begin{array}{l}x>\frac{1}{4}\\ x<\frac{1}{9}\end{array}\right.\Rightarrow$không tìm được x

TH2: $\left\{\begin{array}{l}2x-\frac{1}{2}<0\\ 3x-\frac{1}{3}>0\end{array}\right.\iff \left\{\begin{array}{l}2x<\frac{1}{2}\\ 3x>\frac{1}{3}\end{array}\right.\iff \left\{\begin{array}{l}x<\frac{1}{4}\\ x>\frac{1}{9}\end{array}\right.\iff \frac{1}{9}<x<\frac{1}{4}$

Vậy: $\frac{1}{9}<x<\frac{1}{4}$