Đáp án:
Ta có: `x^2 −2xy+6y^2 −12x+12y+41=0`
`⇔ x^2 −2xy+y^2 −2.6.x+2.6.y+36)+(5y^2 −10y+5)=0`
`⇔(y+6−x)^2 + 5(y−1)^2 =0`
`⇔` $\left \{ {{y + 6 − x = 0} \atop {y − 1 = 0 }} \right.$
`⇔` $\left \{ {{x=7} \atop {y=1}} \right.$
Thay `x=7` ,` y=1` vào A ta được:
`A=(2020−2019(9−x−y)^{2019}−(x−6y)^{2019})/(y^{1010})`
`= (2020−2019(9−7−1)^{2019}−(7−6.1)^{2019})/(1^{1010})`
`= 2020−2019−1=0`
Vậy `A=0`