f(x)=(1/3)x^3-(1/2)ax^2+（a-1)x+1f'(x)=x^2-ax+a-11.当a=0时f'(x)=x^2-1设切点为(x0,y0)a=0f(x)=(1/3)x^3-x+1f'(x)=x^2-1k=y'=x0^2-1k=(y0-7)/(x0-3)y0=1/3*x0^3-x0+1((1/3)x0^3-x0+1-7)/(x0-3)=2x0^2-1解得x0=...