Tìm x:

- Hướng dẫn giải

$\frac{x+\frac{3}{2}}{x-\frac{2}{3}}<0 Đk: x\ne \frac{2}{3}$

Ta có: $x-\frac{2}{3}<x+\frac{3}{2}$ nên

$\left\{\begin{array}{l}x-\frac{2}{3}<0\\ x+\frac{3}{2}>0\end{array}\right.\iff \left\{\begin{array}{l}x<\frac{2}{3}\\ x>-\frac{3}{2}\end{array}\right.\iff -\frac{3}{2}<x<\frac{2}{3}$

Vậy: $-\frac{3}{2}<x<\frac{2}{3}$