m+3>2;2m-4<0解得 -1<m<2|m-2|-|1+m|解:慎闭因为-1<m<2所以m-2<0:1+m>0即|m-2|-|1+m|= -(配孝m-2)培孝稿-(1+m)=-m+2-1-m=1-2m
M最小为0-M-2-1+M=-3
因为m+3>2,2m-4<0,所以-1<m<2,所以原式为2-m-(1+m)=2-m-1-m=1-2m
-1所以,原式等于2-m-1-m=1-2m