kx的平方-x+1+k>0,解不等式
问题描述:
kx的平方-x+1+k>0,解不等式
答
∵kx²-x+1+k>0 ∴(x+1) [kx-(k+1)]>0
(1)当(k+1)/k>﹣1即k>0或k<﹣1/2时, x<﹣1或x>(k+1)/k
(2)当(k+1)/k<﹣1即﹣1/2<k<0时, x>﹣1或x<(k+1)/k