Đặt `A = 2x^2 + 8x + y^2 - 10y + 43`

`= (2x^2 + 8x + 8) + (y^2 - 10y + 25) + 10`

`= 2(x^2 + 4x + 4) + (y^2 - 10y + 25) + 10`

`= 2(x + 2)^2 + (y - 5)^2 + 10`

Vì `2(x+2)^2 >= 0 AA x; (y - 5)^2 >= 0 AA y`

`=> 2(x+2)^2 + (y-5)^2 + 10 >= 10 AA x;y`

Dấu "`=`" xảy ra khi:

`{((x+2)^2 = 0),((y-5)^2 = 0):} <=> {(x+2=0),(y-5=0):} <=> {(x = -2),(y = 5):}`

Vậy `A_(min) = 10 <=> x = -2; y = 5`

`\ttcolor{cyan}{#PUyen}`