可得：k²-3k+1=0 即：k²+1=3k
两边同时除以k得：k+1/k=3
(2k²-5k-2)+3/(k²+1)
=[2(k²+1)-2-5k-2)+3/3k
=[6k-5k-4]+1/k
=k+1/k-4
=3-4
=-1