1. Let $a, b, c$ be real numbers, $abc \neq 0$ and $a+b=$ $c$. Then the value of $\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+$ $\frac{a^{2}+b^{2}-c^{2}}{2 a b}$ is ( ).
(A) -1
(B) 1
(C) 2
(D) 3
$-1 .(\mathrm{B})$.
$$
\begin{aligned}
\text { Original expression }= & \frac{(b-c)^{2}+2 b c-a^{2}}{2 b c} \\
& +\frac{(c-a)^{2}+2 c a-b^{2}}{2 c a} \\
& +\frac{(a+b)^{2}-2 a b-c^{2}}{2 a b} \\
= & \frac{2 b c}{2 b c}+\frac{2 c a}{2 c a}-\frac{2 a b}{2 a b}=1 .
\end{aligned}
$$

Another solution: Take $a=b=1, c=2$ and substitute to get.