1. Solve the inequality $\left(x^{2}-3 x+3\right)^{4 x^{3}+5 x^{2}} \leq\left(x^{2}-3 x+3\right)^{2 x^{3}+18 x}$.
Answer: $x \in\left(-\infty ;-\frac{9}{2}\right] \cup[0 ; 1] \cup\{2\}$.

Solution. The base of the exponent is positive for all $x$, so the given inequality is equivalent to the following:

$$
\left(x^{2}-3 x+2\right)\left(\left(4 x^{3}+5 x^{2}\right)-\left(2 x^{3}+18 x\right)\right) \leq 0 \Leftrightarrow(x-1)(x-2) x\left(2 x^{2}+5 x-18\right) \leq 0 \Leftrightarrow(x-1)(x-2)^{2} x(2 x+9) \leq 0
$$

from which, using the method of intervals, we find that $x \in\left(-\infty ;-\frac{9}{2}\right] \cup[0 ; 1] \cup\{2\}$.