1A. Solve the inequality

$$
(\sqrt{2}+1)^{\frac{|x|-1}{x+1}} \leq(\sqrt{2}-1)^{-\frac{x}{6}}
$$
Solution. Since $(\sqrt{2}-1)^{-1}=\frac{1}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1$, the inequality takes the form $(\sqrt{2}+1)^{\frac{|x|-1}{x+1}} \leq(\sqrt{2}+1)^{\frac{x}{6}}$. From this, $\frac{|x|-1}{x+1} \leq \frac{x}{6}$, i.e., $\frac{6|x|-6-x^{2}-x}{6(x+1)} \leq 0$.

If $x \geq 0$, then $\frac{5 x-6-x^{2}}{6(x+1)} \leq 0$ and the solution is $x \in[0,2] \cup[3, \infty)$.

If $x<0$, then $\frac{-7 x-6-x^{2}}{6(x+1)} \leq 0$ and the solution is $x \in[-6,-1] \cup(-1,0)$.

Finally, $x \in[-6,-1) \cup(-1,2] \cup[3, \infty)$.