当y=kx+2k+1>0
(1) k<0时,kx>-2k-1x<-(2k+1)/k
∵-1≤x≤1
∴-1<-(2k+1)/k (2k+1-k)/k<0 (k+1)/k<0
故-1(2) k=0时,1>0成立 (2)
(3) k>0时,x>-(2k+1)/k
∵-1≤x≤1
∴-(2k+1)/k<1(2k+1-k)/k>0 (k+1)/k>0
k<-1或k>0
故 k>0(3)
综上：k>-1
希望能帮到你,祝学习进步O(∩_∩)O