`x²+y²+xy+3x-3y+9=0`
⇒`4x²+4y²+4xy+12x-12y+36=0`
⇒`(2x)²+2.2x(y+3)+4y²-12y+36=0`
⇒`(2x+y+3)²-(y+3)²+4y²-12y+36=0`
⇒`(2x+y+3)²-y²-6y-9+4y²-12y+36=0`
⇒`(2x+y+3)²+3y²-18y+27=0`
⇒`3(2x+y+3)²+(3y)²-2.3y.9+81=0`
⇒`3(2x+y+3)²+(3y-9)²=0`
Ta có `3(2x+y+3)²+(3y-9)²≥0`
⇒`2x+y+3=0 , 3y-9=0`
⇒`x=-3 , y=3`
Thay `x=-3 , y=3` vào `Q` ta đc
`Q=2`