依基本不等式得
k^2+2/k=k^2+1/k+1/k≥3(k^2·1/k·1/k)^(1/3)=3
→-(k^2+2/k)≤-3
→15/16-(k^2+2/k)≤-33/16
→1/[15/16-(k²+2/k)]≥-16/33
→3/(15/16-2/k-k^2)≥-16/11.
故原式只存在最小值:-16/11.
此时,k^2=1/k,即整数k=1.
依基本不等式得
k^2+2/k=k^2+1/k+1/k≥3(k^2·1/k·1/k)^(1/3)=3
→-(k^2+2/k)≤-3
→15/16-(k^2+2/k)≤-33/16
→1/[15/16-(k²+2/k)]≥-16/33
→3/(15/16-2/k-k^2)≥-16/11.
故原式只存在最小值:-16/11.
此时,k^2=1/k,即整数k=1.