1.46 If $a b c=1$. Then the value of $\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1}$ is
(A) 1.
(B) 0.
(C) -1.
(D) -2.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] Given $a b c=1$, then $a \neq 0, b \neq 0, c \neq 0$, thus
$$
\begin{aligned}
& \frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1} \\
= & \frac{a b c}{a b^{2} c+a b c+b c}+\frac{b}{b c+b+1}+\frac{b c}{a b c+b c+b} \\
= & \frac{1}{b+1+b c}+\frac{b}{b c+b+1}+\frac{b c}{1+b c+b} \\
= & \frac{1+b+b c}{b c+b+1}=1 .
\end{aligned}
$$

Therefore, the answer is $(A)$.