已知a,b为正数,ab-3a-2b=0,求2a+3b的最小值和ab的最小值,求(a-1)(b-1)的最小值
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已知a,b为正数,ab-3a-2b=0,求2a+3b的最小值和ab的最小值,求(a-1)(b-1)的最小值
已知a,b为正数,ab-3a-2b=0,求2a+3b的最小值和ab的最小值,求(a-1)(b-1)的最小值
已知a,b为正数,ab-3a-2b=0,求2a+3b的最小值和ab的最小值,求(a-1)(b-1)的最小值
b=3a/(a-2),因a>0,b>0,故a-2>0,2a+3b=2a+9a/(a-2)=2a+18/(a-2)+9=2(a-2)+18/(a-2)+13>=2√2(a-2)*18/(a-2)+13=25,即
2a+3b的最小值为25.
ab=3a+2b=3a+6a/(a-2)=3a+12/(a-2)+6=3(a-2)+12/(a-2)+12>=2√3(a-2)*12/(a-2)+12=24,ab的最小值为24.
(a-1)(b-1)=ab-(a+b)+1=3a+2b-(a+b)+1=2a+b+1=2a+3a/(a-2)+1=2a+6/(a-2)+4=2(a-2)+6/(a-2)+8>=2√2(a-2)*6/(a-2)+8=4√3+8,即(a-1)(b-1)的最小值为4√3+8.
(1)a>0、b>0.
ab-3a-2b=0→2/a+3/b=1.
故依Cauchy不等式得
(2a+3b)(2/a+3/b)≥(2+3)²
→2a+3b≥25.
故所求最小值为:25.
(2)依基本不等式得
ab=3a+2b≥2√(6ab)
→(ab)²-24ab≥0.
显然,ab>0,故
ab≥...
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(1)a>0、b>0.
ab-3a-2b=0→2/a+3/b=1.
故依Cauchy不等式得
(2a+3b)(2/a+3/b)≥(2+3)²
→2a+3b≥25.
故所求最小值为:25.
(2)依基本不等式得
ab=3a+2b≥2√(6ab)
→(ab)²-24ab≥0.
显然,ab>0,故
ab≥24,
即所求最小值为:24.
(3)(a-1)(b-1)
=ab-a-b+1
=(3a+2b)-a-b+1
=2a+b+1
=2[2b/(b-3)]+b+1
=(b-3)+12/(b-3)+8
≥2√[(b-3)·12/(b-3)]+8
=4√3+8
即所求最小值为:4√3+8.
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