3B. Solve the inequality:

$$
\frac{1}{2^{2 x}+3} \geq \frac{1}{2^{x+2}-1}
$$
Solution. The left side of the inequality is defined and positive for every real number $x$, while the right side is defined for every number $x \neq -2$, and the expression can take negative values. Then

1) If the right side is negative, i.e., if $2^{x+2}-2 < 0$, then the equation is equivalent to the inequality $2^{2 x}+3 \leq 2^{x+2}-1$, or the inequality $2^{2 x}-4 \cdot 2^{x}+4 \leq 0$. We make the substitution $2^{2 x}=t$, which transforms the inequality into $t^{2}-4 t+4 \leq 0$ or $(t-2)^{2} \leq 0$. Its only solution is $t=2$, or $x=1$.

Finally, the solution to the original inequality is $x \in (-\infty, -2) \cup \{1\}$.

## IV year