`5x^2+5y^2+8xy-2x+2y+2=0`
`<=> (4x^2+8xy+4y^2)+(x^2-2x+1)+(y^2+2y+1)=0`
`<=> 4(x+y)^2+(x-1)^2+(y+1)^2=0`
Do `(x+y)^2>=0; (x-1)^2>=0; (y+1)^2>=0` với `AAx;y`
`=> {(x+y=0),(x-1=0),(y+1=0):}`
`<=> {(x=1),(y=-1):}`
Thay `x=1;y=-1` vào M ta có:
`M=(1-1)^2019+(1-2)^2020+(-1+1)^2021`
`M=0^2019+(-1)^2020+0^2021`
`M=1`
Vậy `M=1`