1. Let $a, b$ be rational numbers, and $|a|>0$, the equation $||x-a|-b|=5$ has three distinct roots. Then $b=(\quad)$.
(A) 1
(B) 3
(C) 5
(D) 6
-,1. C.
From the given, we have
$$
|x \sim a|=b+5 \text { or } b-5 \text {. }
$$

And each equation has two, one, or zero solutions.
Since the original equation has three distinct roots, one equation has two solutions, and the other has one solution.
Therefore, one of the non-negative numbers $b+5, b-5$ must be 0.
Thus, $b=5$.
At this point, $x=a+10, a-10, a$ are all solutions to the original equation.