Question

己知关于x的分式方程k/x+1+x+k/x-1=1的解为负数，求K的取值范围

Answer 1

k/(x+1)+(x+k)/(x-1)=1,
k/(x+1)+(k+1)/(x-1)=0,
两边都乘以(x+1)(x-1),得kx-k+(k+1)x+k+1=0,
(2k+1)x=-1,
-1≠x=-1/(2k+1)<0,
∴2k+1>0,2k+1≠1，
∴k>-1/2,且k≠0.

Answer 2

解：去分母得k(x-1)+(x+k)(x+1)=(x+1)(x-1)
kx-k+x²+kx+x+k=x²-1
(2k+1)x=－1
x=－1/(2k+1)
因为x为负数，所以－1/(2k+1)<0，
解得k>－1/2