原式=[(x+k/x(x-1)]-(x-1/3x)=x/3(x-3)
3(x+k)-(x-1)^2=x^2
3x+3k-2x^2+2x-1=0
∵此方程有增根,
∴x=1和0
即当x=1时
3+3k-2+2-1=0
3k=-2
k=-2/3
即当x=0时
3k-1=0
3k=1
k=1/3
原式=[(x+k/x(x-1)]-(x-1/3x)=x/3(x-3)
3(x+k)-(x-1)^2=x^2
3x+3k-2x^2+2x-1=0
∵此方程有增根,
∴x=1和0
即当x=1时
3+3k-2+2-1=0
3k=-2
k=-2/3
即当x=0时
3k-1=0
3k=1
k=1/3