首先[6(3k-1)]^2-4(k^2-1)*72＞0
得（k-3)^2＞0得k≠3,
又(k^2-1)≠0得k≠±1
因?yàn)閤1+x2=72/(x^2-1)
若兩根為整數(shù),(k^2-1)為72的因數(shù),而72的因數(shù)只能為1,2,3,4,6,8,9,12,18,24,36,72
所以k^2-1=1,2,3,4,6,8,9,12,18,24,36,72
k=±√2,±√3,±√4,±√5,±√7,±√9,±√10,±√13,±√19,±√25,±√37,±√73
只討論k=±√4,±√9,±√25就可以了
所以k=±2,-3,-5
k為2,-2,-3,-5時(shí),方程(k^-1)x^-6(3k-1)x+72等于0有兩個(gè)不相等的正整數(shù)根