当k为何值时,分式方程k-2/(x-2)=k+[1/(x-2)]无解.

两边乘x-2
k-2=k(x-2)+1
k-2=kx-2k+1
kx=3k-3
k=0时,是0=0-3,不成立,无解
k≠0,x=(3k-3)/k,无解则这是增根,即分母为0
x-2=0
x=2
所以(3k-3)/k=2
3k-3=2k
k=3
所以k=0,k=3