x²-x+1=(x-1/2)²+3/4>0
∴原不等式可化為：
0<3x^2+kx+6≤6(x^2-x+1)
①3x^2+kx+6>0
顯然Δ=k²-72<0
②3x^2+kx+6≤6(x^2-x+1)
,可化解為3x²-（6+k）x≥0
則有Δ=（6+k）²≤0
解兩個(gè)不等式得：k=-6
即k=-6時(shí),不等式0<(3x^2+kx+6)/(x^2-x+1)≤6對(duì)任意實(shí)數(shù)x恒成立