f(x)=1/3*x³-mx²-3m²x+1当m=1时,f(x)=1/3*x³-x²-3x+1当x=2时,f(2)=1/3*8-4-6+1=-19/3f'(x)=x²-2x-3,那么f'(2)=4-4-3=-3所以f(x)在点(2,f(2))处的切线方程为：y-(-19/3)=-3(x-2),整理...