关于x的方程(k+1)x²+(3k-1)x+2k-2=0?
△=（3k-1)^2-4(k+1)(2k-2)
=9k^2-6k+1-8k^2+8
=k^2-6k+9
=(k-3)^2,
(2)x=[1-3k+k-3]/[2(k+1)]=-1,或x=(2-2k)/(k+1)=-2+4/(k+1)为整数,
∴(k+1)是4的约数,
∴正整数k=1或3.