${ x^2-2xy+3y-5x+7=0}$
⇔ ${x^2-(2xy+5x)+\dfrac{(2y+5)^2}{4}-\dfrac{(2y+5)^2}{4}+(3y+7)=0}$
⇔ ${\bigg[ x^2-x.(2y+5)+\bigg(\dfrac{(2y+5)}{2}\bigg)^2\bigg]-\dfrac{(2y+5)^2}{4}+(3y+7)=0}$
⇔ ${\bigg[x^2-\dfrac{2.x.(2y+5)}{2}+\bigg(\dfrac{(2y+5)}{2}\bigg)^2\bigg]-\dfrac{4y^2+20y+25}{4}+\dfrac{4.(3y+7)}{4}=0}$
⇔ ${\bigg(x-\dfrac{2y+5}{2}\bigg)^2-\dfrac{4y^2+20y+25-4.(3y+7)}{4}=0}$
⇔ ${\bigg(x-\dfrac{2y+5}{2}\bigg)^2-\dfrac{4y^2+20y+25-12y-28}{4}=0}$
⇔ ${\bigg(x-\dfrac{2y+5}{2}\bigg)^2-\dfrac{4y^2+8y-3}{4}=0}$

⇔ ${\bigg(x-\dfrac{2y+5}{2}\bigg)^2-\dfrac{4.(y^2+2y+1)-7}{4}=0}$
⇔ ${\bigg(x-\dfrac{2y+5}{2}\bigg)^2-\dfrac{4.(y+1)^2-7}{4}=0}$
⇔ ${\bigg(x-\dfrac{2y+5}{2}\bigg)^2-(y+1)^2+\dfrac{7}{4}=0}$
⇔ ${\bigg(x-\dfrac{2y+5}{2}\bigg)^2-(y+1)^2=\dfrac{-7}{4}}$

⇔ ${\bigg[\bigg(x-\dfrac{2y+5}{2}-y-1\bigg).\bigg(x-\dfrac{2y+5}{2}+y+1\bigg)\bigg]=\dfrac{-7}{4}}$
⇔ ${\bigg(\dfrac{2x}{2}-\dfrac{2y}{2}-\dfrac{5}{2}-\dfrac{2y}{2}-\dfrac{2}{2}\bigg).\bigg(\dfrac{2x}{2}-\dfrac{2y}{2}-\dfrac{5}{2}+\dfrac{2y}{2}+\dfrac{2}{2}\bigg)=\dfrac{-7}{4}}$

⇔ ${\dfrac{2x-2y-5-2y-2}{2}.\dfrac{2x-2y-5+2y+2}{2}=\dfrac{-7}{4}}$
⇔ ${\dfrac{2x-4y-7}{2}.\dfrac{2x-3}{2}=\dfrac{-7}{4}}$
⇔ ${\dfrac{(2x-4y-7).(2x-3)}{4}=\dfrac{-7}{4}}$
⇔ ${(2x-4y-7).(2x-3)=-7 (1)}$
$\text{Vì x, y ∈ Z nên 2x-4y-7 ∈ Z và 2x-3 ∈ Z (2)}$ 
$\text{ Từ (1) và (2) suy ra : (2x-4y-7).(2x-3)=-7=1.(-7)=-7.1=-1.7=7.(-1)}$

$\text{ +) Trường hợp 1}$

$\left \{ {{2x-4y-7=-7} \atop {2x-3=1}} \right.$

⇔ $\left \{ {{2x-4y=0} \atop {2x=4}} \right.$

⇔ $\left \{ {{4-4y=0} \atop {x=2}} \right.$

⇔ $\left \{ {{4y=4} \atop {x=2}} \right.$

⇔ $\left \{ {{y=1} \atop {x=2}} \right.$

$\text{ +) Trường hợp 2}$

$\left \{ {{2x-4y-7=1} \atop {2x-3=-7}} \right.$

⇔ $\left \{ {{2x-4y=8} \atop {2x=-4}} \right.$

⇔ $\left \{ {{-4-4y=8} \atop {x=-2}} \right.$

⇔ $\left \{ {{4y=-12} \atop {x=-2}} \right.$

⇔ $\left \{ {{y=-3} \atop {x=-2}} \right.$

$\text{ +) Trường hợp 3}$

$\left \{ {{2x-4y-7=-1} \atop {2x-3=7}} \right.$

⇔ $\left \{ {{2x-4y=6} \atop {2x=10}} \right.$

⇔ $\left \{ {{10-4y=6} \atop {x=5}} \right.$

⇔ $\left \{ {{4y=4} \atop {x=5}} \right.$

⇔ $\left \{ {{y=1} \atop {x=5}} \right.$

$\text{ +) Trường hợp 4}$

⇔ $\left \{ {{2x-4y=14} \atop {2x=2}} \right.$

⇔ $\left \{ {{2-4y=14} \atop {x=1}} \right.$

⇔ $\left \{ {{4y=-12} \atop {x=1}} \right.$

⇔ $\left \{ {{y=-3} \atop {x=1}} \right.$

$\text{ Vậy các nghiệm nguyên (x;y) của phương trình là: (-2;-3),(2;1),(5;1),(1;-3)}$

Giải thích các bước giải:

`x^2 - 2xy+3y−5x+7=0`

`⇔4x^2−8xy+12y−20x+28=0` (nhân với `4`)

`⇔(4x^2−6x)−(8xy−12y)−(14x−21)+7=0`

`⇔2x(2x−3)−4y(2x−3)−7(2x−3)+7=0`

`⇔(2x−3)(2x−4y−7)=−7`

`x;y∈Z` và `7` là `SNT`

`⇒` Có `4` Trường Hợp sau:

`-` Trường Hợp `1`:

 `2x−3=−1`

`2x−4y−7=7`

`⇔x=1 ; y=−3`

`-` Trường Hợp `2`: 

`2x−3=1`

`2x−4y−7=−7`

`⇔x=2;y=1`

`-`  Trường Hợp `3`:

`2x−3=−7`

`2x−4y−7=1`

`⇔x=−2;y=−3`

`-` Trường Hợp `4`:

`2x−3=7 ⇔x=5`

`2x−4y−7=−1 <=> y=4`