一元二次方程x²+2x+a-1=0有两个根为x1,x2,且x1-x1x2=0∵x1+x2=-2∴x2=-2-x1∵x1-x1x2=0∴x1+x1(2+x1)=0x1²+3x1=0x1(x1+3)=0x1=0或x1=-3当x1=0时0+0+a-1=0a=1当x1=-3时9-6+a-1=0∴a=-2综上a=1或-2...