Solution. Let's introduce the substitution $t=2012^{x}$. We solve the inequality

$$
t^{2}-t-2011 \cdot 2012 \geq 0 \quad \text { or } \quad(t-2012) \cdot(t+2011) \geq 0
$$

Its solution is

$$
t \leq-2011 \quad \text { or } \quad t \geq 2012
$$

Thus,

$$
2012^{x} \leq-2011 \quad \text { or } \quad 2012^{x} \geq 2012
$$

The first inequality is impossible, and from the second we get $x \geq 1$.
Task B-3.1. Solve the inequality

$$
2012^{2 x}-2012^{x}-2011 \cdot 2012 \geq 0
$$