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

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

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

from which, using the method of intervals, we find that $x \in(-\infty ;-2] \cup\{-1\} \cup\left[0 ; \frac{5}{2}\right]$.