tìm gtnn

H

huan2122000

1, A= \frac{ x^2 - 2x + 2010}{ x^2 } = 1 - \frac{ 2 }{ x } + \frac{ 2010 }{ x^2 }
đặt \frac{ 1 }{ x } = y
\Rightarrow A= 2010y^2 - 2y + 1
= 2010 ( y^2 - \frac{ y }{ 1005 } + \frac{ 1 }{ 2010 } )
=2010 ( y^2 - \frac{ 2y }{ 2010 } + \frac{ 1 }{ 2010^2 } + \frac{ 1 }{ 2010 } - \frac{ 1 }{ 2010^2 } )
=2010 ( y - \frac{ 1 }{ 2010 } ) ^2 + \frac{ 2009}{ 2010 } \geq \frac{ 2009 }{ 2010 }
dau = xay ra khi y - \frac{ 1 }{ 2010 } = 0\Rightarrow y = \frac{ 1 }{ 2010 } \Rightarrow \frac{ 1 }{ x } = \frac{ 1 }{ 2010 } \Rightarrow x = 2010
:p:p:p:p:p:p:p:p
 
V

vermouthvinyard

A= $ \frac{x^2 -2x + 2010}{x^2} $
= $ 1-\frac{2x}{x^2} + \frac{2010}{x^2} $
= $ 1- \frac{2}{x}+\frac{2010}{x^2} $
Đặt $ t= \frac{1}{x} $
=> A= $ 2010t^2 - 2t + 1 $
= $ 2010( t^2 - \frac{1}{1005}t ) + 1 $
= $ 2010( t- \frac{1}{2010}) ^2 + 1- \frac{1}{2010^2} $
= $ 2010 ( t- \frac{1}{2010}) ^2 + \frac{2009}{2010} $ \geq$ \frac{2009}{2010} $
$vậy MinA= \frac{2009}{2010} $ \Leftrightarrow t= $ \frac{1}{2010} $
\Rightarrow $ x= 2010 $
 
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