1. Given real numbers $x, y, z$ satisfy $x+y+z=0$ and $x y z>0$. Let $T=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, then ().
(A) $T>0$
(B) $T=0$
(C) $T<0$
(D) The sign of $T$ cannot be determined
$$
-1 . C .
$$

Since $x y z>0$, then $x, y, z$ are all non-zero.
$$
\begin{array}{l}
\text { Therefore, } x^{2}+y^{2}+z^{2}>0 . \\
\text { Also, }(x+y+z)^{2} \\
=x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=0,
\end{array}
$$

Thus, $x y+y z+z x<0$.
$$
\text { Therefore, } T=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{x y+y z+z x}{x y z}<0 \text {. }
$$