题160911（14分）已知三次函数$fleft( x ight)$有三个零点${{x}_{1}},{{x}_{2}},{{x}_{3}}$，且在点$left( {{x}_{i}},fleft( {{x}_{i}} ight) ight)$处切线的斜率为${{k}_{i}}$（$i=1,2,3$）．
证明：$frac{1}{{{k}_{1}}}+frac{1}{{{k}_{2}}}+frac{1}{{{k}_{3}}}=0$．
证：由题意不妨设$fleft( x ight)=aleft( x-{{x}_{1}} ight)left( x-{{x}_{2}} ight)left( x-{{x}_{3}} ight)$（$a e 0$），则
$f'left( x ight)=aleft[ left( x-{{x}_{2}} ight)left( x-{{x}_{3}} ight)+left( x-{{x}_{1}} ight)left( x-{{x}_{3}} ight)+left( x-{{x}_{1}} ight)left( x-{{x}_{2}} ight) ight]$
故$dfrac{1}{{{k}_{1}}}+dfrac{1}{{{k}_{2}}}+dfrac{1}{{{k}_{3}}}$
$=dfrac{1}{f'left( {{x}_{1}} ight)}+dfrac{1}{f'left( {{x}_{2}} ight)}+dfrac{1}{f'left( {{x}_{3}} ight)}$
$=dfrac{1}{aleft( {{x}_{1}}-{{x}_{2}} ight)left( {{x}_{1}}-{{x}_{3}} ight)}+dfrac{1}{aleft( {{x}_{2}}-{{x}_{1}} ight)left( {{x}_{2}}-{{x}_{3}} ight)}+dfrac{1}{aleft( {{x}_{3}}-{{x}_{1}} ight)left( {{x}_{3}}-{{x}_{2}} ight)}$
$=dfrac{left( {{x}_{2}}-{{x}_{3}} ight)+left( {{x}_{3}}-{{x}_{1}} ight)+left( {{x}_{1}}-{{x}_{2}} ight)}{aleft( {{x}_{1}}-{{x}_{2}} ight)left( {{x}_{1}}-{{x}_{3}} ight)left( {{x}_{2}}-{{x}_{3}} ight)}$
$=0$