若实数a,b满足ab-4a-b+1=0(a>1),则(a+1)(b+2)的最小值为?答案：27

Question

若实数a,b满足ab-4a-b+1=0(a>1),则(a+1)(b+2)的最小值为?
答案：27

Answer 1

ab+b+2a+2
ab-4a-b+1==0
所以
ab+ab-4a+1+2a+2==2ab-2a+3
b==(4a-1)/(a-1) (a>1)
===4+3/(a-1)
原式===6a+6a/(a-1)+3
===6a+6/(a-1)+9
==6(a-1)+6/(a-1)+15
在a==2时最小值为 12+15===27