1.21 Real numbers $x, y, z$ satisfy $x+y+z=x y z$, then the following statements I. If two of $x, y, z$ are opposite numbers, then the third number must be 0. II. If one of $x, y, z$ is 0, then the other two must be opposite numbers.
III. $z=\frac{x+y}{x y-1}$.
(A) Only I is true, II and III are not true.
(B) Only I is not true, II and III are true.
(C) I, II, and III are all not true.
(D) I, II, and III are all true.
(China Beijing High School Mathematics Competition, 1991)
[Solution] (1) If $x+y=0$, and $x=y=0$, then $z=0$. If $x \neq 0$, then $|x|=|y| \neq 0$, by the problem statement we have $z=xyz$. Since $xy \neq 1$ (x, y have opposite signs), from $(1-xy)z=0$ we get $z=0$. Therefore, (I) is true.
(2) Suppose $x=0$, then $xyz=0$, hence $y+z=0$. Therefore, (II) is true.
(3) If $xy-1=0$, then $xy=1$, meaning $x, y$ have the same sign, so $x+y \neq 0$. Therefore, $x+y=xyz-z=(xy-1)z$ does not hold.
Thus, $xy-1 \neq 0$. Hence, $z=\frac{x+y}{xy-1}$ always holds. In summary, (I), (II), and (III) are all true. Therefore, the answer is $(D)$.