=[(x+1)(x+4)+2]/(x+1)
=(x+1)(x+4)/(x+1)+2/(x+1)
=x+4+2/(x+1)
=(x+1)+2/(x+1)+3
x>-1
x+1>0
所以(x+1)+2/(x+1)>=2√[(x+1)*2/(x+1)]=2√2
所以1/y>=3+2√2
所以0
函數(shù)y=(x+1)/(x^2+5x+6) (x>-1) 的最大值
函數(shù)y=(x+1)/(x^2+5x+6) (x>-1) 的最大值
基本不等式的應(yīng)用,
基本不等式的應(yīng)用,
數(shù)學(xué)人氣：567 ℃時(shí)間：2020-06-18 07:08:14
優(yōu)質(zhì)解答
1/y=(x²+5x+6)/(x+1)
=[(x+1)(x+4)+2]/(x+1)
=(x+1)(x+4)/(x+1)+2/(x+1)
=x+4+2/(x+1)
=(x+1)+2/(x+1)+3
x>-1
x+1>0
所以(x+1)+2/(x+1)>=2√[(x+1)*2/(x+1)]=2√2
所以1/y>=3+2√2
所以0所以最大值=3-2√2
=[(x+1)(x+4)+2]/(x+1)
=(x+1)(x+4)/(x+1)+2/(x+1)
=x+4+2/(x+1)
=(x+1)+2/(x+1)+3
x>-1
x+1>0
所以(x+1)+2/(x+1)>=2√[(x+1)*2/(x+1)]=2√2
所以1/y>=3+2√2
所以0
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