設(shè)k為整數(shù),且k不等于零,方程kx^2-(k-1)x+1=0有有理數(shù)根,求k的值
△=(k-1)^2-4*k*1
=k^2-2k+1-4k
=k^2-6k+1>0
k^2-6k+9+1-9>0
(k-3)^2-8>0
(k-3)^2>8
k-3=2√2
k=2√2+3