已知关于x的方程x2-2（k+1）x+k2+2k-1=0①（1）试判断方程①的根的情况；（2）如果a是关于y的方程y2-（x1+x2-2k）y+（x1-k）（x2-k）=0②的根，其中x1，x2为方程

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