解:集合B包含于 集合A ,分两种情况 :①B为空集时:有:m+1>2m+1 解得:m<0②B不为空集时,满足:2m+1>=m+1,得:m>=0m+1>=3,得:m>=22m+1<=6,得:m<=5/2取交集得:2<=m<=5/2综合①②,取并集得:m的取值范围为:m<0或2<=m<=5/2