9.065. $\frac{15}{4+3 x-x^{2}}>1$.

9.065. $\frac{15}{4+3 x-x^{2}}>1$.
## Solution.

Let's move 1 to the left side of the inequality and bring it to a common denominator. We have $\frac{x^{2}-3 x+11}{x^{2}-3 x-4} - 1 = \frac{x^{2}-3 x+11 - (x^{2}-3 x-4)}{x^{2}-3 x-4} = \frac{15}{x^{2}-3 x-4}$. Since $x^{2}-3 x-4 \neq 0$ for $x \in R$, then $x^{2}-3 x-4 < 0$. Solving this inequality, we find $-1 < x < 4$.

Answer: $\quad x \in (-1 ; 4)$