若函数f(x)=|x+m-1|/(x-2),(m>0),且f(1)=-1,求实数k的取值范围,使得方程f(x)=kx有且

f(1) = |m|/(-1) = -1 ,m>0 ,m=1
f(x) = kx |x|/(x-2) = kx
k*x^2 - 2*kx - |x| =0
x>=0 ,x[kx - (2k+1)] = 0
x< 0 ,x[kx + (1 - 2k)] = 0
当k=0 ,只有 x=0 一个解
当 k不等于0 x=0 ,x= (2k+1)/k ,x= (2k-1)/k
(1) 若有3解
x = (2k+1)/k >0 k< -1/2 或 k>0 1）
x = (2k-1)/k