Đáp án:
`A=13x^2+y^2+4xy-2y-16x+2019`
`=4x^2+9x^2+y^2+4xy-2y-4x-12x+2019`
`=(4x^2+4xy+y^2-4x-2y+1)+(9x^2-12x+4)+2014`
`=(2x+y-1)^2+(3x-2)^2+2014`
Ta có: `(2x+y-1)^2>=0; (3x-2)^2>=0`
`=> (2x+y-1)^2+(3x-2)^2>=0`
`=> (2x+y-1)^2+(3x-2)^2+2014>=2014`
Dấu "=" xảy ra `<=> ` $\left\{\begin{matrix}2x+y-1=0& \\3x-2=0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x=\dfrac{2}{3}& \\y=-\dfrac{1}{3}& \end{matrix}\right.$
Vậy `A_(min)=2014 <=>` $\left\{\begin{matrix}x=\dfrac{2}{3}& \\y=-\dfrac{1}{3}& \end{matrix}\right.$
