第1个回答 2014-12-28
解:
由4a+b+5=ab得:
b(a-1)=4a+5
显然a≠1,故有
b=(4a+5)/(a-1)
则ab=a(4a+5)/(a-1)=a(4a-4+9)/(a-1)=4a+9a/(a-1)=4a+9(a-1+1)/(a-1)
=4a+9+9/(a-1)
=4(a-1)+9/(a-1)+13
若0<a<1,
ab=4(a-1)+9/(a-1)+13<=-2√4(a-1)·9/(a-1)+13=1
当且仅当4(a-1)=9/(a-1), 即a=-1/2时,"="成立,与a>0不符
若a>1
ab=4(a-1)+9/(a-1)+13≥2√4(a-1)·9/(a-1)+13=25
当且仅当4(a-1)=9/(a-1), 即a=5/2时,"="成立