Violympic

B

baochauhn1999

Ta có:
$x+2y=5$
$<=>x=5-2y$
Như vậy:
$A=x^2+y^2=(5-2y)^2+y^2=4y^2+25-20y+y^2=5y^2-20y+25=5(y^2-4y+5)=5(y^2-4y+4+1)=5(y-2)^2+5$\geq$5$

$"=" <=> y=2;x=1$
 
H

huynhbachkhoa23

$x+2y=5$ [TEX]\Rightarrow[/TEX] $x=5-2y$
$A=(2y-5)^2+y^2=5y^2-20y+25=5(y-2)^2+5$
$A_{min}=5$ [TEX]\Leftrightarrow[/TEX] $x=1,y=2$
 
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