Question

当k取何值时,关于x的方程(k^2-1)x ^2+(3k-9)x=18的根都是整数

Answer 1

①k²-1=0,则k=±1

即：(3k-9)x=18

k=1时,x=-3,

k=-1时,x=-3/2,舍

②k²-1=0,则k≠±1

(k²-1)x²+(3k-9)x=18

(k-1)(k+1)x²+(3k-9)x-18=0

即：[(k-1)x-3][(k+1)x+6]=0

x1=3/(k-1),x2=6/(k+1)

3/(k-1)是整数,则k=2,0,-2,4

6/(k+1)是整数,则k=5,2,1(舍),0,-2,-3,-4,-7

找出共同的

∴k=2,0,-2

综上,k=1或2或0或-2