1. Given that $x, y, z$ are not all equal real numbers, and $x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}=k$. Then all possible values of $k$ are ( ).
(A) $\pm 1$
(B) $\pm \frac{1}{2}$
(C) $\pm 2$
(D) cannot be determined
-、1.A.
Given $x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}=k$, we have
$$
\begin{array}{l}
\frac{1}{x}=\frac{y}{k y-1}, \\
z=\frac{1}{k-y} .
\end{array}
$$

Substituting equations (1) and (2) into $z+\frac{1}{x}=k$, we get
$$
\begin{array}{l}
\frac{1}{k-y}+\frac{y}{k y-1}=k \\
\Rightarrow k y-1+y(k-y)=k(k-y)(k y-1) \\
\Rightarrow k^{3} y-k^{2}-k^{2} y^{2}+1-k y+y^{2}=0 \\
\Rightarrow k y\left(k^{2}-1\right)-\left(k^{2}-1\right)-y^{2}\left(k^{2}-1\right)=0 \\
\Rightarrow\left(k^{2}-1\right)\left(k y-1-y^{2}\right)=0 .
\end{array}
$$

Therefore, $k^{2}-1=0$ or $k y-1-y^{2}=0$, i.e., $k= \pm 1$ or $k=y+\frac{1}{y}$.
Substituting $k=y+\frac{1}{y}$ into the given equations, we get $x=y=$ $z$, which contradicts the given conditions, while $k= \pm 1$ is possible.