1. The positions of the three numbers $a$, $b$, and $c$ on the number line are shown in the figure, and $|b|=|c|$. Then, simplify $\sqrt{b^{2}}-|a-b|+|a-c|-|b+c|=$ $\qquad$ .
二、1.3c or $-3 b$.
$\because a 、 b$ are negative numbers, $c$ is a positive number, and $|a|>|b|=|c|$,
$$
\begin{aligned}
\therefore \text { original expression } & =-b-(b-a)+c-a \\
& =-2 b+c=3 c \text { or }-3 b .
\end{aligned}
$$