a>=0已有人证了,
f`(x)=a/(ax+1)+3x^2-2x-a,令f'(x)>=0,因x>=1,ax+1>0,不等式两边消去一个x,并同乘以ax+1,
整理得(3x-2)(ax+1)>=a^2,因y=3x-2,y=ax+1在[1,+∞)上是增函数,所以有
1×(a+1)>=a^2,解得:(1-根号5)/2<=a<=(1+根号5)/2
故0<=a<=<=(1+根号5)/2
已知f(x)＝ln(ax+1)+x^3-x^2-ax
=0,因x>=1,ax+1>0,不等式两边消去一个x,并同乘以ax"}}}'>