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Albert
Well-known member
- Jan 25, 2013
- 1,225
Given:
$x,y>0$ and $9x+4y=2005$, find:
$\min\left(\dfrac{1}{x}+\dfrac{1}{y}\right)$
$x,y>0$ and $9x+4y=2005$, find:
$\min\left(\dfrac{1}{x}+\dfrac{1}{y}\right)$
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