己知关于x的分式方程k⼀x+1+x+k⼀x-1=1的解为负数,求K的取值范围

2025-02-24 05:49:18
推荐回答(2个)
回答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.

回答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