$-1 \leq \frac{x^{2}-5 x+6}{x^{2}-9} \leq 1$ and $x^{2}-3 x+2>0, \neq 1$ $\frac{(x-3)(2 x+1)}{x^{2}-9} \geq 0 \mid \frac{5(x-3)}{x^{2}-9} \geq 0$ The solution to this inequality is $x \in\left[\frac{-1}{2}, \infty\right)-\{3\}$ for $x^{2}-3 x+2>0$ and $\neq 1$ $x \in(-\infty, 1) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\}$ Combining the two solution sets (taking intersection) $x \in\left[-\frac{1}{2}, 1\right) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\}$