`2x²+10y²-6xy-6x-2y+10=0`
⇒`4x²+20y²-12xy-12x-6y+20=0`
⇒`(2x)²-2.2x(3y+3)+20y²-6y+20=0`
⇒`(2x-3y-3)²-(3y+3)²+20y²-6y+20=0`
⇒`(2x-3y-3)²-9y²-18y-9+20y²-6y+20=0`
⇒`(2x-3y-3)²+11y²-24y+11=0`
⇒`11(2x-3y-3)²+(11y)²-2.11y.11+11²=0`
⇒`11(2x-3y-3)²+(11y-11)²=0`
Ta có `11(2x-3y-3)²+(11y-11)² ≥ 0 ∀x,y`
Dấu `=` xảy ra khi `2x-3y-3 = 0 , 11y-11=0`
⇒ `x=3 , y=1`
Thay `x=3 , y=1` vào A ta đc
A=`-1/3`