1. If real numbers $a, b, c$ satisfy $|a|+a=0,|a b|=a b,|c|$ $-c=0$, then the algebraic expression $\sqrt{b^{2}}-|a+b|-\sqrt{c^{2}-2 b c+b^{2}}$ $+|a-c|$ simplifies to ( ).
(A) $2 c-b$ or $b$
(B) $2 c-b$
(C) $b$
(D) -6
$-1 .(A)$.
$$
\because|a|=-a, \therefore a \leqslant 0 \text {. }
$$

If $a<0$, from the problem we know $b \leqslant 0, c \geqslant 0$.
$\therefore$ the original expression $=-b+a+b-(c-b)-(a-c)=b$;
If $a=0$, from the problem we know $b \geqslant 0$ or $b<0, c \geqslant 0$.
Then when $b<0$, it is easy to know the original expression $=b$;
When $b \geqslant 0$, the original expression $=b-b \pm(c-b)+c=2 c-b$ or
b.