9.080. $\frac{x^{4}-2 x^{2}-8}{x^{2}+2 x+1}<0$.

9.080. $\frac{x^{4}-2 x^{2}-8}{x^{2}+2 x+1}<0$.
Solution.

Domain of definition: $\quad x \neq-1$.

By solving the biquadratic equation $x^{4}-2 x^{2}-8=0$, we can represent the inequality as $\frac{\left(x^{2}+2\right)\left(x^{2}-4\right)}{(x+1)^{2}}0$ for $x \in R$ and
$(x+1)^{2}>0$ for $x \neq-1$, then this inequality is equivalent to the system of inequalities $\left\{\begin{array}{l}(x-2)(x+2)<0, \\ x \neq-1 .\end{array}\right.$

Answer: $\quad x \in(-2 ;-1) \cup(-1 ; 2)$.