A function is given by the following formula, where $x$ is the variable, and $a$ and $b$ are constants, neither of which is equal to 1. What could be the domain of the function? What is its range?

$$
\frac{x}{1-x}+\frac{a}{(1-x) \cdot(1-a)}+\frac{b-x}{(1-x) \cdot(1-a) \cdot(1-b)}
$$
The domain of the function includes all numbers except 1, $a$, and $b$, where the function is undefined.

To determine the range, we can simplify the function as follows:

$$
y=\frac{1}{1-x}\left(x+\frac{a}{1-a}+\frac{b-x}{(1-a)(1-b)}\right)
$$

Then, we add the fractions inside the large parentheses. A suitable common denominator is the denominator of the third fraction, so by extending the fractions, the sum of the numerators is

$$
\begin{gathered}
x(1-a)(1-b)+a(1-b)+b-x= \\
=(a+b-a b)-x(a+b-a b)=(1-x)(a+b-a b)
\end{gathered}
$$

(we expanded and then factored the sum back into a product). Noticing the possibility of simplification,

$$
y=\frac{1}{1-x} \cdot \frac{(1-x)(a+b-a b)}{(1-a)(1-b)}=\frac{a+b-a b}{(1-a)(1-b)}
$$

(since $1-x \neq 0$, as $x=1$ is not in the domain). Thus, the value of $y$ is independent of $x$ and is the number on the right-hand side, which alone constitutes the range of our function.

Pintér János (Székesfehérvár, Teleki B. Gymnasium II. grade)