k取什么整数值时,下列方程有整数解? （k²-1）x²-6（3k-1）x+72=0

(k+1)(k-1)x^2-(18k-6)+6*12=0
[(k+1)x-12] *[(k-1)x-6]=0
若k-1,则有：x1=12/(k+1),此时,k+1为12的因数,即k+1=1,2,3,4,6,12,-1,-2.,-3.-4,-6,-12
故k=0.1,2,3,5,11,-2,-3.-4.-5,.-7,-13
若k1,则有：x2=6/(k-1),此时k-1为6的因数,即：k-1=1,2,3,6,-1.-2.-3.-6
故k=2,3,4,7,0.-1.-2,-5
若k1,-1,则有两个解x1,x2
综合得：只要k=0,1,2,3,4,5,7,11,-1.-2.-3.-4.-5.-7.-13,则原方程都有整数解.