p:|1-(x-1)/3|=|(4-x)/3|<=2∴|x-4|<=6-6<=x-4<=6,即-2<=x<=10q:x²-2x 1<=m²∴(x-1)²<=m²-m<=x-1<=m,即1-m<=x<=1 mp是q的充分不必要条件,即-2<=x<=10在1-m<=x<=1 m上1-m<=-2,1 m>=10∴m>=9综上,m范围为[9, ∞)
/1-x-1/3|小于等于2?
