（1）
kx^2-2(k+1)x+k-1=0有實數(shù)根,則
①k=0時,-2x-1=0 得 x=-1/2滿足
①k≠0時,
Δ=[-2(k+1)]^2-4k(k-1)≥0
4k^2+8k+4-4k^2+4k≥0
12k≥-4
k≥-1/3
綜上,所求k的取值范圍為k≥-1/3
（2）
由（1）得,方程有兩個實數(shù)根x1,x2(x1＜x2),則k>-1/3且k≠0.
x1+x2=2(k+1)/k x1x2=(k-1)/k
(x2-x1)²=(x1+x2)²-4x1x2
=[2(k+1)/k]²-4(k-1)/k
=(4k²+8k+4)/k²-(4k²-4k)/k²
=(12k+4)/k²
x2-x1=√[(12k+4)/k²]=√(12k+4)/k
∴y=kx2-kx1=k(x2-x1)=k*√(12k+4)/k=√(12k+4)=2√(3k+1) (k>-1/3且k≠0)
即這個函數(shù)的解析為y=2√(3k+1) (k>-1/3且k≠0)???????????????k??????????????????????x2-x1=??[(12k+4)/k²]=-/+????(12k+4)/k?? ???????????????y=2??(3k+1)??y=-2??(3k+1)??????????????????????????????? (x2-x1)²=(12k+4)/k²(k>-1/3??k??0)??-1/30???x2-x1=??[(12k+4)/k²]=??(12k+4)/ky=kx2-kx1=k(x2-x1)=k*??(12k+4)/k=??(12k+4)=2??(3k+1) ???????????????? {-2??(3k+1)??-1/30??