What is the value of the sum $A+B$, if

$$
\frac{1}{A}=1-\frac{1-x}{y}, \quad \frac{1}{B}=1-\frac{y}{1-x}, \quad x=\frac{1-a}{1-\frac{1}{a}}, \quad y=1-\frac{1}{x}
$$

and $a \neq \pm 1$.
According to the third relationship, $x$ depends on $a$, and according to the fourth relationship, $y$ depends on $x$, so ultimately the value of $A+B$ is determined by the value of $a$ according to the first two relationships:

$$
\frac{1}{A}=\frac{x+y-1}{y}, \quad \frac{1}{B}=\frac{1-x-y}{1-x}=\frac{x+y-1}{x-1}
$$

The numerators of the two expressions are identical. Thus,

$$
A+B=\frac{y+(x-1)}{x+y-1}=1
$$

However, we must determine the conditions under which our result is valid. $A$, and $B$ are defined if

$$
\begin{gathered}
y \neq 0, \text { and } \frac{1}{A} \neq 0 \text {, that is, } \frac{1-x}{y} \neq 1 \text {, or } 1-x \neq y, \quad x+y-1 \neq 0 \\
1-x \neq 0 \text {, and } \frac{1}{B} \neq 0 \text {, that is, } \frac{y}{1-x} \neq 1 \text {, or } x \neq 1 \text {, and } x+y-1 \neq 0
\end{gathered}
$$

It will suffice to examine the last two conditions, because if these are satisfied, the division in (1) can also be performed, and from the fourth relationship, $y=0$ cannot result.

$y$ is not defined in the case of $x=0$, but this is excluded by $a \neq 1$. $x$ is not defined, beyond the case of $a=1$, also in the case of $a=0$, which we must exclude.

$x=1$ would only occur in the case of $a=1 / a$, which is excluded by $a \neq \pm 1$. Finally, the condition $x+y-1 \neq 0$ transforms, considering the fourth relationship, then the third relationship, and the previous considerations, as follows:

$$
x-\frac{1}{x} \neq 0, \quad x^{2} \neq 1, \quad x \neq-1, \quad a-1 \neq 1-\frac{1}{a}
$$

$a^{2}-2 a+1=(a-1)^{2} \neq 0$, which is again satisfied.

According to all this, $A+B=1$ if $a \neq-1,0,1$.