`-3x^2-2xy+4x-4y=1`

`<=>` `-3x^2+4x-2y(x+2)=1`

`<=>` `2y(x+2)=-3x^2+4x-1`

TH1: `x+2=0=>0=-21=>` Vô lý

TH2: `x+2 ne 0`

`2y(x+2)=-3x^2+4x-1<=>2y=(-3x^2+4x-1)/(x+2)`

`=(-3x^2-6x+10x+20-21)/(x+2)`

`=(-3x(x+2)+10(x+2)-21)/(x+2)`

`=10-3x-21/(x+2)`

Để `2y in ZZ` thì `(10-3x-21/(x+2)) in ZZ`

Mà `(10-3x) in ZZ` do `x in ZZ => 21/(x+2) in ZZ`

Do `x in ZZ => x+2 in ZZ => x+2` là ước của `21`

`=>` `(x+2) in {-21;-7;-3;-1;1;3;7;21}`

`=>` `x in {-23;-9;-5;-3;-1;1;5;19}`

TH1: `x=-23 => 2y=80=>y=40 in ZZ  (tm)`

TH2: `x=-9=>2y=40=>y=20 in ZZ  (tm)`

TH3: `x=-5=>2y=32=>y=16 in ZZ  (tm)`

TH4: `x=-3=>2y=40=>y=20 in ZZ  (tm)`

TH5: `x=-1=>2y=-8=>y=-4 in ZZ  (tm)`

TH6: `x=1=>2y=0=>y=0 in ZZ  (tm)`

TH7: `x=5=>2y=-8=>y=-4 in ZZ  (tm)`

TH8: `x=19=>2y=-48=>y=-24 in ZZ  (tm)`

Vậy `(x;y) in {(-23;40);(-9;20);(-5;16);(-3;20);(-1;-4);(1;0);(5;-4);(19;-24)}` thỏa mãn bài toán