问题描述: 若实数a.b满足ab-4a-b+1=o(a>1),求实数(a+1)*(b+2)的最小值 1个回答 分类:数学 2014-12-02 问题解答: 我来补答 解;由a>1可得a-1>0因为ab-4a-b+1=0(a>1)所以b=(4a-1)/(a-1)设f(a)=(a+1)(b+2)将b=(4a-1)/(a-1)代入可得f(a)=(a+1)(b+2)=(a+1)((4a-1)/(a-1)+2)=(6a-3)(a+1)/(a-1)=6(a-1)+[6/(a-1)]+15由于a-1>0所以1/(a-1)>0由均值不等式可知:6(a-1)+[6/(a-1)]≥2*√(6(a-1)*[6/(a-1)]=12取等条件是1/(a-1)=a-1那么a-1=1,得a=2所以f(a)=(a+1)(b+2)≥12+15=27所以最小值为:27 展开全文阅读