1.求y=(x^2+x+1)/(x^2+2x+1),(x>0)的最小值.2.当x>3时,求y=2x^2/(x-3)的最小值 3.当0

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1.求y=(x^2+x+1)/(x^2+2x+1),(x>0)的最小值.2.当x>3时,求y=2x^2/(x-3)的最小值 3.当0
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1.求y=(x^2+x+1)/(x^2+2x+1),(x>0)的最小值.2.当x>3时,求y=2x^2/(x-3)的最小值 3.当0
1.求y=(x^2+x+1)/(x^2+2x+1),(x>0)的最小值.
2.当x>3时,求y=2x^2/(x-3)的最小值
3.当0

1.求y=(x^2+x+1)/(x^2+2x+1),(x>0)的最小值.2.当x>3时,求y=2x^2/(x-3)的最小值 3.当0
1.y=(x^2+x+1)/(x^2+2x+1)
=(x^2+2x+1-x)/(x^2+2x+1)
=1-x/(x^2+2x+1)
-x/(x^2+2x+1)上下同除以x,
=1-1/(x+2+1/x)
因为x>0,用均值不等式即可解决.
2.y=2x^2/(x-3)
=[2(x-3)^2+12(x-3)+18]\(x-3)
=2(x-3)+12+18/(x-3)
做法同题1
3.y=(a^2/x)+(b^2/(1-x)
=[(a^2/x)+(b^2/(1-x)]*[x+(1-x)]
拆开,同上,均值不等式
第3题,百分百对!

一:
y=((x+1)^2-x)/(x+1)^2=1-x/(x+1)^2
令y'=x/(x+1)^2 (求其最大值,x>0)
y'=1/(x+2+1/x)<=1/(2+2*(x*1/x)^(1/2))=1/4
所以y的最小值为1-1/4=3/4
二:
y=2((x-3)^2+6*x-9)/(x-3)=2*(x-3)+12+18/(x-3)
令...

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一:
y=((x+1)^2-x)/(x+1)^2=1-x/(x+1)^2
令y'=x/(x+1)^2 (求其最大值,x>0)
y'=1/(x+2+1/x)<=1/(2+2*(x*1/x)^(1/2))=1/4
所以y的最小值为1-1/4=3/4
二:
y=2((x-3)^2+6*x-9)/(x-3)=2*(x-3)+12+18/(x-3)
令t=x-3>0
y=2*t+18/t+12>=2*((2*t*18/t)^(1/2))+12=24
所以y的最小值为24
三:
y>=2(a^2*b^2/(x-x^2))^1/2=2*|a*b|*(1/(x-x^2))^1/2 (1)
令y'=-x^2+x<=1/4 代入(1)
y>=2|a*b|*2=4*|ab|

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