$a$) `y(x-2) + 3(x-6) = 2`
`⇔ y(x-2) + 3(x-2) - 12 = 2`
`⇔ (y+3)(x-2) = 14`
`⇒` `y+3;x-2` `∈` `Ư(14)={±1;±2;±7;±14}`
Ta có bảng:
$\left[\begin{array}{ccc}x-2&-14&-7&-2&-1&1&2&7&14\\y+3&-1&-2&-7&-14&14&7&2&1\\x&-12&-5&0&1&3&4&9&16\\y&-4&-6&-10&-17&11&4&-1&-2\end{array}\right]$
Vậy `(x;y)=(-12;-4);(-5;-6);(0;-10);(1;-17);(3;11);(4;4);(9;-1);(16;-2)`
$b$) `xy + 3x - 2y - 7 = 0`
`⇔ x(3+y) - 2y - 7 = 0`
`⇔ x(3+y) - 2y - 6 = 1`
`⇔ x(3+y) - 2(y + 3) = 1`
`⇔ (x-2)(y+3) = 1`
`⇒` `x-2;y+3` `∈` `Ư(1)={±1}`
Ta có bảng:
$\left[\begin{array}{ccc}x-2&-1&1\\y+3&-1&1\\x&1&3\\y&-4&-2\end{array}\right]$
Vậy `(x;y)=(1;-4);(3;-2)`
$c$) `xy - x + 5y - 7 = 0`
`⇔ x(y-1) + 5y - 7 = 0`
`⇔ x(y-1) + 5y - 5 = 2`
`⇔ x(y-1) + 5(y-1) = 2`
`⇔ (x+5)(y-1) = 2`
`⇒` `x+5;y-1` `∈` `Ư(2)={±1;±2}`
Ta có bảng:
$\left[\begin{array}{ccc}x+5&-2&-1&1&2\\y-1&-1&-2&2&1\\x&-7&-6&-4&-3\\y&0&-1&3&2\end{array}\right]$
Vậy `(x;y)=(-7;0);(-6;-1);(-4;3);(-3;2)`
