1. If $x, y, z$ are all real numbers, and satisfy
$$
\begin{array}{l}
\frac{x y}{(y+z)(z+x)}+\frac{y z}{(z+x)(x+y)}+ \\
\frac{z x}{(x+y)(y+z)}=1,
\end{array}
$$

then the values of $x, y, z$ are ( ).
(A) all positive
(B) all non-negative
(C) all negative
(D) exactly one is zero
$-1 . \mathrm{D}$.
Obviously, $(x+y)(y+z)(z+x) \neq 0$.
Eliminating the denominator, we get
$$
\begin{array}{l}
x y(x+y)+y z(y+z)+z x(z+x) \\
=(x+y)(y+z)(z+x) .
\end{array}
$$

Simplifying and organizing, we get $x y z=0$.
Therefore, at least one of $x, y, z$ is zero.
If two or three of $x, y, z$ are zero, then at least one of $x+y, y+z, z+x$ is zero, making the equation meaningless.

Hence, exactly one of $x, y, z$ is zero.