\[\frac{(2x + 3)(3x - 4)}{x - 1} \leq 0\]
\[\frac{6 \cdot (x + 1,5)\left( x - 1\frac{1}{3} \right)}{x - 1} \leq 0\]
\[Ответ:( - \infty; - 1,5\rbrack \cup \left( 1;1\frac{1}{3} \right\rbrack.\]