(k-1)/[(x+1)(x-1)]-1/[x(x-1)]=(k-5)/[x(x+1)]
[x(k-1)-(x+1)]/[x(x-1)]-[(k-5)(x-1)]/[x(x+1)]=0
[3x+k-6]/[x(x+1)(x-1)]=0
[3x+k-6]=0
k=6-3x=6+3=9→(当x=-1时) ∴选D. 9.
看不清的次数都是二次
(k-1)/[(x+1)(x-1)]-1/[x(x-1)]=(k-5)/[x(x+1)]
[x(k-1)-(x+1)]/[x(x-1)]-[(k-5)(x-1)]/[x(x+1)]=0
[3x+k-6]/[x(x+1)(x-1)]=0
[3x+k-6]=0
k=6-3x=6+3=9→(当x=-1时) ∴选D. 9.