Solve the equation $\left[\frac{1}{1-x}\right]=\left[\frac{1}{1.5-x}\right]$ on the set of real numbers. ([a], the integer part of $a$ is the greatest integer not greater than $a$.)
Let's examine the values of the two fractions first between -1 and 0:

$$
-1 \leq \frac{1}{1-x} < 0 = \left[\frac{1}{1.5-x}\right]$, there is no solution here.

Similarly, if $\frac{1}{2} \leq x < 1$, then $\left[\frac{1}{1-x}\right] = 2$ and $\left[\frac{1}{1.5-x}\right] = 1$, which also has no solution.

If $x \geq 1$, then $\left[\frac{1}{1-x}\right] = -1$ and $\left[\frac{1}{1.5-x}\right] = 0$, which has no solution either.

If $x < 0$, then $\left[\frac{1}{1-x}\right] = 0$ and $\left[\frac{1}{1.5-x}\right] = -1$, which is a solution.

If $x \geq 2.5$, then $\left[\frac{1}{1-x}\right] = -1$ and $\left[\frac{1}{1.5-x}\right] = -2$, which is also a solution.

For the case where $0 \leq x < \frac{1}{2}$, let $\left[\frac{1}{1-x}\right] = k$, where $k$ is an integer. The first inequality rearranges to $x \geq \frac{1}{k} + 0.5$, so $k \geq 2$. The second inequality is: $x < \frac{1}{k} + 1.5$, which means $\frac{1}{k(k-1)} > \frac{1}{2}$, or $k(k-1) < 2$, which is impossible.

Therefore, the solution to the problem is: $x < 0$, or $x \geq 2.5$.

Note. Some solvers read the solutions from the graphs of the functions $\left[\frac{1}{1-x}\right]$ and $\left[\frac{1}{1.5-x}\right]$.