f'(x)=3x^2-3a
設(shè)過點(diǎn)(-2,1)的y=f(x)切線的切點(diǎn)橫坐標(biāo)為x
則切線斜率為3x^2-3a
所以(x^3-3ax+3-1)/(x+2)=3x^2-3a
x^3-3ax+2=3x^3-3ax+6x^2-6a
2x^3+6x^2-6a-2=0
x^3+3x^2-3a-1=0
根據(jù)題意,上述一元三次方程需有三個(gè)不相同的實(shí)數(shù)根
判別式：A=9 B=9(3a+1)C=9(3a+1)
△=B^2-4AC<0
[9(3a+1)]^2-4*9*[9(3a+1)]<0
(3a+1)^2-4(3a+1)<0
(3a+1)(3a-3)<0
(3a+1)(a-1)<0
-1/3