由数轴可知 a﹤b﹤0﹤c
|a|﹥c ,a+c﹤0
c-b﹥0
a﹤0,b﹤0 ,b+a﹤0
原式=|a+c|+|c-b|-|b+a|
=-(a+c)+c-b+b+a
=-a-c+c-b+b+a
=0