## Task B-1.1.

Solve the inequality in the set of real numbers

$$
\frac{1}{1-\frac{1}{1+\frac{1}{x-1}}} \leqslant 2019
$$
## Solution.

The condition for the solution of the given inequality is $x-1 \neq 0$ and $1+\frac{1}{x-1} \neq 0,1$, that is, $x \neq 0$ and $x \neq 1$.

By simplifying the left side of the inequality, we sequentially obtain

$$
\begin{array}{ll}
\frac{1}{1-\frac{1}{\frac{x}{x-1}}} \leqslant 2019 . & 1 \text{ point} \\
\frac{1}{1-\frac{x-1}{x}} \leqslant 2019 . & \\
\frac{1}{\frac{x-x+1}{x}} \leqslant 2019 . & 1 \text{ point } \\
&
\end{array}
$$

Thus, $x \leqslant 2019$.

Finally, the solution to the given inequality is $x \in (-\infty, 2019] \setminus \{0,1\}$.