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

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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)的最小值若实数a.b满足ab-4a-

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
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

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
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

由ab-4a-b+1=0.===>b=(4a-1)/(a-1).===>b+1=(5a-2)/(a-1).===>(a+1)(b+1)=13+5(a-1)+[6/(a-1)].由a>1及均值不等式可知,(a+1)(b+1)≥13+2√30.等号仅当a=(5+√30)/5,b=(24+4√30)/30时取得。故[(a+1)(b+1)]min=13+2√30.