a,b,c均为正数,且点(a+b+c,c)在直线ax+by=3上,则2a+b+c的最小值为A.-1  

a,b,c均为正数,且点(a+b+c,c)在直线ax+by=3上,则2a+b+c的最小值为

A.-1                 B.+1              C.-2             D.+2

答案

答案:

C  ∵点(a+b+c,c)在直线上,

∴a(a+b+c)+bc=3=(-1)2.

∴(a+b)(a+c)=(-1)2,

而2a+b+c=(a+b)+(a+c)≥

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