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pinkcoi

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thaiha_98

$x^2 + 2y^2 + 2xy – 2x – 6y +10$
$=x^2 + 2xy + y^2 -2(x+y) + 1 + y^2 - 4y + 4+6$
$=(x+y)^2 - 2(x+y) + 1 + (y-2)^2 + 6$
$=(x+y-1)^2 + (y-2)^2 + 6$
Ta có: $(x+y-1)^2 \ge 0; (y-2)^2 \ge 0$
\Rightarrow $Min(x^2 + 2y^2 + 2xy – 2x – 6y +10)=6$
Dấu "=" xảy ra \Leftrightarrow $x= -1 ; y=2$
 
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