Question

k是整数,(k^2+2k-3)/(9-k^2)也是整数,求k的值

Answer 1

原式=(k+3)(k-1)/[-(k+3)(k-3)]

=-(k-1)/(k-3)

=-[(k-3+2)/(k-3)]

=-1-2/(k-3)是整数

所以k-3=±1,±3

且分母9-k²≠0

所以k=2,k=4,k=0,k=6