解由2+lx-1l-lx+3l=0得lx+3l-lx-1l=2当x≥1时，lx+3l-lx-1l=x+3-（x-1）=4≠2当x≤-3时，lx+3l-lx-1l=-x-3-[-（x-1）]=-x-3+x-1=-4≠2当-3＜x＜1时，lx+3l-lx-1l=2变为x+3-[-(x-1)]=2即x+3+x-1=2解得x=0故方程2+lx-1l-lx+3l=0的解的个数为1个选A
