∵由图可得:c﹤a﹤0﹤b
∴a-c﹥0,a-b﹤0 ,b-c﹥0,2a﹤0
∴丨a-c丨-丨a-b丨-丨b-c丨+丨2a▏=a-c-[-﹙a-b﹚]-﹙b-c﹚﹢﹙﹣2a﹚
=a-c+a-b-b+c﹣2a
=-2