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
因为x1+x2=72/(x^2-1)
若两根为整数,(k^2-1)为72的因数,而72的因数只能为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
只讨论k=±√4,±√9,±√25就可以了
所以k=±2,-3,-5
k为2,-2,-3,-5时,方程(k^-1)x^-6(3k-1)x+72等于0有两个不相等的正整数根