`5x^2 +5y^2 +2x+4xy-4y+10`
`= (x^2 +4xy+4y^2 )+4x^2 +4x-2x -4y+10`
`=[(x+2y)^2 -2(x+2y)+1]+(4x^2 +4x+1) +8`
`=(x+2y-1)^2 +(2x+1)^2 +8`
Vậy `Min=8`
Khi
$\left \{ {{}(x+2y-1)^2 =0 \atop {(2x+1)^2 =0}} \right.$
$\Rightarrow \left \{ {{x+2y-1=0} \atop {2x+1=0}} \right.$
$\Rightarrow \left \{ {{x+2y-1=0} \atop {2x=-1}} \right.$
$\Rightarrow \left \{ {{x+2y=1} \atop {x=- \dfrac{1}{2}}} \right.$
$\Rightarrow \left \{ {{- \dfrac{1}{2}+2y=1} \atop {x=- \dfrac{1}{2}}} \right.$
$\Rightarrow \left \{ {{2y=\dfrac{3}{2}} \atop {x=- \dfrac{1}{2}}} \right.$
$\Rightarrow \left \{ {{y=\dfrac{3}{4}} \atop {x=- \dfrac{1}{2}}} \right.$
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