$a$) $x(y-2) = -12$
$⇒$ $x;y-2$ $∈$ `Ư(12)={±1;±2;±3;±4;±6;±12}`
Ta có bảng:$\left[\begin{array}{}x&-12&-6&-4&-3&-2&-1&1&2&3&4&6&12\\y-2&1&2&3&4&6&12&-12&-6&-4&-3&-2&-1\\y&3&4&5&6&8&14&-10&-4&-2&-1&0&1\end{array}\right]$
Vậy `(x;y)=(-12;3);(-6;4);(-4;5);(-3;6);(-2;8);(-1;14);(1;-10);(2;-4);(3;-2);(4;-1);(6;0);(12;1)`
$b$) $xy - 3x - y = 0$
$⇔ x(y-3) - y + 3 = 3$
$⇔x(y-3) - (y-3) = 3$
$⇔ (x-1)(y-3)=3$
$⇒ x-1;y-3$ $∈$ `Ư(3)={±1;±3}`
Ta có bảng :
$\left[\begin{array}{ccc}x-1&-3&-1&1&3\\y-3&-1&-3&3&1\\x&-2&0&2&4\\y&2&0&6&4\end{array}\right]$
Vậy `(x;y)=(-2;2);(0;0);(2;6);(4;4)`
$c) xy + 2x + 2y = -10$
$⇔ x(2+y) + 2y + 4 = -6$
$⇔ x(2+y) + 2(2+y) = -6$
$⇔ (x+2).(y+2) = -6$
$⇒ x+2;y+2$ $∈$ `Ư(6)={±1;±2;±3;±6}`
Ta có bảng:
$\left[\begin{array}{ccc}x+2&-6&-3&-2&-1&1&2&3&6\\y+2&1&2&3&6&-6&-3&-2&-1\\x&-8&-5&-4&-3&-1&0&1&4\\y&-1&0&1&4&-8&-5&-4&-3\end{array}\right]$
Vậy `(x;y)=(-8;-1);(-5;0);(-4;1);(-3;4);(-1;-8);(0;-5);(1;-4);(4;-3)`
