（m+1)x2﹣（m+1）x+m≤0解集为实数集R
(1)m+1<0
(2)判别式＝（m+1)^2-4m(m+1)<=0
(1)解得m<-1
(2)解得：(m+1)(m+1-4m)<=0
(m+1)(-3m+1)<=0
(m+1)(3m-1)>=0
m>=1/3或m<=-1
当m+1=0,即m=-1时,不等式是-1<=0,恒成立
综上所述,m=<-1