答：
f(k)=4k+ 1/(2k+3)²
f(k)=(2k+3)+(2k+3)+1/(2k+3)² -6
>=3³√[(2k+3)*(2k+3)*1/(2k+3)²]-6
=3-6
=-3
當(dāng)且僅當(dāng)2k+3=1/(2k+3)²即2k+3=1即k=-1時取得最小值
因為：k>0
所以：f(k)是k的單調(diào)遞增函數(shù)
所以：f(k)不存在最大值,也不存在最小值答案是五分之一f(k)=(4k+1) /(2k+3)²
=(4k+6-5)/(2k+3)²
=2 /(2k+3) -5/(2k+3)² 設(shè)t=1/(2k+3)
=-5t²+2t
=-5*(t²-2t/5+1/25)+1/5
=-5(t-1/5)²+1/5
當(dāng)t=1/5時取得最小值1/5
此時t=1/(2k+3)=1/5，k=1

所以：k=1時取得最小值1/5