`5x^2+2y^2+2xy-26x-16y+54`
`=(x^2+2xy+y^2)-(10x+10y)+(4x^2-16x+16)+(y^2-6y+9)+29`
`=(x+y)^2-10(x+y)+25+(2x-4)^2+(y-3)^2+4`
`=(x+y-5)^2+(2x-4)^2+(y-3)^2+4>=4`
Dấu = xảy ra khi `{(x+y=5),(2x-4=0),(y-3=0):}`
`<=> {(x=2),(y=3):}`
Vậy `min_(5x^2+2y^2+2xy-26x-16y+54)=4<=>(x;y)=(2;3)`