已知k是方程x^2-100x+1=0的一个不为0的根,不解方程,你能求出k^2-99k+100\k^2+1的值吗?如果能

将k代入x^2-100x+1=0得：
k^2-100k+1=0
k^2+1=100k
(k+1/k)=100
k^2-99k+100/(k^2+1)
=k^2-99k-k+k+1-1+100/(k^2+1)
=(k^2-100k+1)+k-1+100/(100k)
=0-1+k+1/k
=-1+(k+1/k)
=-1+100
=99