30. Find the number of natural numbers $n$, not exceeding 3134, for which the equation $x^{[x]}=n$ has a solution. Here $[x]$ is the greatest integer not exceeding $x$.
Answer: 422.

III. Solve the inequality

$$
\frac{4^{-|x-2|}}{\sqrt{x^{2}-x-2}+2} \leqslant \frac{2^{1-|x|}}{\sqrt{x^{2}+6 x}+4}.
$$

In the answer, write the smallest root by absolute value, rounding it to two decimal places if necessary. If there are no solutions, then write the number 0.

Solution. Rewrite the inequality as

$$
\frac{4^{-|x-2|}}{\sqrt{(x+1)(x-2)}+2} \leqslant \frac{2^{-|x|}}{\sqrt{\left(\frac{x}{2}+3\right)\left(\frac{x}{2}\right)}+2}
$$

If we denote the left side by $f(x)$, then the right side is $f\left(\frac{x}{2}+2\right)$, and we get the inequality

$$
f(x) \leqslant f\left(\frac{x}{2}+2\right)
$$

Let's analyze the monotonicity. The domain of the original inequality is the set $(-\infty ;-6] \cup[2 ; \infty)$

On the set $x \leqslant-6$, both the left and right sides of the inequality are increasing, so we get $x \leqslant \frac{x}{2}+2$.

On the set $x \geqslant 2$, both the left and right sides of the inequality are decreasing, so we get $x \geqslant \frac{x}{2}+2$.

Since the solution to the inequality will be $x \in(-\infty ;-6] \cup[4 ; \infty)$, then $x=4$.

Answer: 4.