1. 19 If $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}=1$. Then,
(A) at least one of $x, y, z$ is equal to 1.
(B) $x, y, z$ are all equal to 1.
(C) $x, y, z$ are all not equal to 1.
(D) none of the above conclusions is correct.
(8th "Jinyun Cup" Junior High School Mathematics Invitational, 1991)
[Solution] From the given $\frac{1}{x+y+z}=1$, we have $x+y+z=1$.
Then
$x+y=1-z$.

Also, from $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$, we have $\quad x y+y z+z x-x y z=0$.
That is, $x y(1-z)+z(x+y)=0$.
From the $\circledast$ equation, we have $\quad(x y+z)(1-z)=0$.
Substituting $z=1-x-y$ into the first factor, we get

That is,
$$
\begin{array}{c}
(x y+1-x-y)(1-z)=0, \\
(1-x)(1-y)(1-z)=0 .
\end{array}
$$
$\therefore x, y, z$ must have one of them equal to 1.
Therefore, the answer is $(A)$.