依题意,知x1和x2是方程x²-2(k-1)x+k²-7=0的两根
韦达定理：x1+x2=2(k-1)
x1x2=k²-7
因为(x1)²+(x2)²
=(x1+x2)²-2x1x2
=[2(k-1)]²-2(k²-7)
=4(k²-2k+1)-(2k²-14)
=2k²-8k+18
又(x1)²+(x2)²=10
故2k²-8k+18=10
k²-4k+4=0
(k-2)²=0
解得k=2
答案：k=2