Đáp án`+`Giải thích các bước giải:

`x^2-3y^2+2xy-4x-4y+8=0`

`(x^2+2xy+y^2)-4y^2-4x-4y+8=0`

`(x+y)^2-4(x+y)+8-4y^2=0`

`(x+y)^2-2*(x+y)*2+2^2+4-4y^2=0`

`(x+y-2)^2-4y^2+4=0`

`(x+y-2)^2-(2y)^2=-4`

`(x+y-2-2y)(x+y-2+2y)=-4`

`(x-y-2)(x+3y-2)=-4`

`(x-y-2)(x+3y-2)=1*(-4)=(-4)*1=2*(-2)=(-2)*2`

Vì `x,y in ZZ` nên ta có các `TH` sau:

`+) TH1 : {(x-y-2=1),(x+3y-2=-4):}`

`<=> {(x-y=3),(x+3y=-2):}`

`<=> {(3x-3y=9),(x+3y=-2):}`

`<=> {(4x=7),(3x-3y=9):}`

`<=> {(x=7/4 (loại)),(3x-3y=9):}`

`+) TH2 : {(x-y-2=-4),(x+3y-2=1):}`

`<=> {(x-y=-2),(x+3y=3):}`

`<=> {(3x-3y=-6),(x+3y=3):}`

`<=> {(4x=-3),(x+3y=3):}`

`<=> {(x=-3/4 (loại)),(x+3y=-3):}`

`+) TH3 : {(x-y-2=2),(x+3y-2=-2):}`

`<=> {(x-y=4),(x+3y=0):}`

`<=> {(3x-3y=12),(x+3y=0):}`

`<=> {(4x=12),(x+3y=0):}`

`<=> {(x=3 (tm)),(3+3y=0):}`

`<=> {(x=3),(y=-1(tm)):}`

`+) TH4 : {(x-y-2=-2),(x+3y-2=2):}`

`<=> {(x-y=0),(x+3y=4):}`

`<=> {(3x-3y=0),(x+3y=4):}`

`<=> {(4x=4),(x+3y=4):}`

`<=> {(x=1(tm)),(1+3y=4):}`

`<=> {(x=1),(y=1(tm)):}`

` Vậy ` `(x;y) in {(3;-1);(1;1)}`