若解分式方程(k-1)/(x²-1)-1/(x²-x)=(k-5)/(x²+x)时有增根x=-1,则k的值为?

Question

若解分式方程(k-1)/(x²-1)-1/(x²-x)=(k-5)/(x²+x)时有增根x=-1,
则k的值为?

Answer 1

(k-1)/(x²-1)-1/(x²-x)=(k-5)/(x²+x) (k-1)/[(x+1)(x-1)]-1/[x(x-1)]=(k-5)/[x(x+1)] 方程两边同时乘x(x+1)(x-1) (k-1)x-(x+1)=(k-5)(x-1) kx-x-x-1=kx-k-5x+5 kx-x-x-kx+5x=-k+5+1 3x=6-k x=(6-k)/3...