(x-m-1)(x-2m+1)>0
方程(x-m-1)(x-2m+1)=0的两个根是x1=m+1、x2=2m-1
(1)若m=2,则:x1=x2,此时不等式的解集是:{x|x≠2};
(2)若mx2,则解集是:{x|x>m+1或x2,则:x12m-1或x
(x-m-1)(x-2m+1)>0
方程(x-m-1)(x-2m+1)=0的两个根是x1=m+1、x2=2m-1
(1)若m=2,则:x1=x2,此时不等式的解集是:{x|x≠2};
(2)若mx2,则解集是:{x|x>m+1或x2,则:x12m-1或x