kx²+2kx+1=x²+3x-k
(k-1)x²+(2k-3)x+k+1=0
要有实数根,则Δ＞0
(2k-3)²-4(k-1)(k+1)＞0
4k²-12k+9-4k²+4＞0
-12k+13＞0
k＜13/12还有Δ=0吧那么Δ≥0k≤13/12