1.51 Given $a b c=1, a+b+c=2, a^{2}+b^{2}+c^{2}=3$, then the value of $\frac{1}{a b+c-1}$ $+\frac{1}{b c+a-1}+\frac{1}{c a+b-1}$ is
(A) 1.
(B) $\frac{1}{2}$
(C) 2.
(D) $-\frac{2}{3}$.
(China Shanxi Province Taiyuan City Junior High School Mathematics Competition, 1997)
［Solution］From the given conditions, we have
$$
a b+c-1=a b+(2-a-b)-1=(a-1)(b-1) \text {, }
$$

Similarly, $b c+a-1=(b-1)(c-1)$,
$$
\text { and } c a+b-1=(c-1)(a-1) \text {. }
$$

Then $\quad$ the original expression $=\frac{a+b+c-3}{(a-1)(b-1)(c-1)}=\frac{-1}{(a-1)(b-1)(c-1)}$
$$
\begin{array}{l}
=\frac{-1}{a b c-(a b+b c+c a)+a+b+c-1} \\
=\frac{-1}{(a b+b c+c a)+2},
\end{array}
$$

and $a b+b c+c a=\frac{1}{2}\left[(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right]=\frac{1}{2}$. Therefore, $\quad$ the original expression $=-\frac{2}{3}$.
Hence, the answer is $(D)$.