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已知ab=3a+2b,ab都大于0,则ab的最小值为?
人气:369 ℃ 时间:2020-06-03 13:25:29
解答
ab=3a+2b
ab-3a=2b
a=2b/(b-3)

ab=2b²/(b-3)
∵ab>0
∴b-3>0
ab=2b²/(b-3)
=(2(b²-9)+18)/(b-3)
=2(b+3)+18/(b-3)
=2(b-3)+12+18/(b-3)
=2(b-3)+18/(b-3)+12
∵2(b-3)+18/(b-3)≥2√2(b-3)*18/(b-3)=6
∴ab≥6+12=18
即ab最小值为18错的。。唉,确实错了有个地方算错了∵2(b-3)+18/(b-3)≥2√2(b-3)*18/(b-3)=2√36=12∴ab≥12+12=24
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