解:x²-2x-3≤0(x-3)(x+1)≤0-1≤x≤3x²-2mx+m²-4≤0[x-(m+2)][x-(m-2)]≤0m-2≤x≤m+2(1)A∩B=[0,3]m-2=0 m=2(2)A包含于B补,即A∩B=Φm+2<-1或m-2>3m<-3或m>5
A=[-1,3]B=[-2+m,2+m]A∩B=[0,3]-2+m=02+m≥3m=2