已知关于x的方程x²-2(k-1)x+k²=0有两个实数根x1 x2,且/x1+x2/=x1x2-1

x1+x2=2（k-1)x1*x2=k2|2k-1|=k2-12k-1≥0,2k-1=k2-1 k=2,0 舍去0,k=22k-1＜0,1-2k=k2-1 k2+2k-2=0 k=1-√3