已知a>1,ab=2a+ b则(a +1)(b+2)的最小值是

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/14 15:12:02

已知a>1,ab=2a+ b则(a +1)(b+2)的最小值是
已知a>1,ab=2a+ b则(a +1)(b+2)的最小值是

已知a>1,ab=2a+ b则(a +1)(b+2)的最小值是
把ab=2a+ b同除以ab,得到一个关系式
完后把1乘到(a +1)(b+2),再看看
只引导,不告诉答案,有什么疑问再追

我会了,用基本不等式
ab=2a+b≥(2a*b)^0.5
ab≥8
原式=2ab+2≥18

解(a +1)(b+2)
=ab+2a+b+2
=ab+ab+2
=2ab+2
又由a>1,ab=2a+ b
知ab-b=2a
即(a-1)b=2a
即b=2a/(a-1)
故(a +1)(b+2)
=2ab+2
=2a×2a/(a-1)+2
=4a^2/(a-1)+2
=4[(a-1)^2+2a-1]...

全部展开

解(a +1)(b+2)
=ab+2a+b+2
=ab+ab+2
=2ab+2
又由a>1,ab=2a+ b
知ab-b=2a
即(a-1)b=2a
即b=2a/(a-1)
故(a +1)(b+2)
=2ab+2
=2a×2a/(a-1)+2
=4a^2/(a-1)+2
=4[(a-1)^2+2a-1]/(a-1)+2
=4[(a-1)^2+2(a-1)+1]/(a-1)+2
=4[(a-1)+1/(a-1)+2]+2
=4[(a-1)+1/(a-1)]+10
≥4×2√(a-1)*1/(a-1)+10
=8+10
=18

(a +1)(b+2)的最小值是18.

收起