已知三次函数y=f(x)有三个零点x1 x2 x3 且在点(x1,f(x1))处的切线斜率为ki(i=1,2,3),则1/k1+1/k2+1/k3

由题意,f(x)=a(x-x1)(x-x2)(x-x3)则f'(x)=a(x-x2)(x-x3)+a(x-x1)(x-x3)+a(x-x1)(x-x2)令S=a(x1-x2)(x1-x3)(x2-x3)k1=f'(x1)=a(x1-x2)(x1-x3)=S/(x2-x3)k2=f'(x2)=a(x2-x1)(x2-x3)=S/(x3-x1)k3=f'(x3)=a(x3-x1)(x3-x2...