Bài 2: Tìm các số nguyên x, y biết: a) (x – 2) (y + 3) = 7 b) (x + 1) (2y – 3) = 10 c) xy – 3x = -19 d) 3x + 4y – xy = 16

$\text{a) Ta có} :7=1.7=(-1).(-7)$

$\left[\begin{array}{ccc}x-2&1&7&-1&-7\\y+3&7&1&-7&-1\\x&3&9&1&-5\\y&4&-2&-4&-11\end{array}\right]$

$\text{b) Ta có} :10=1.10=(-1).(-10)=2.5=(-2).(-5)$

$\left[\begin{array}{ccc}x+1&1&10&-1&-10&2&5&-2&-5\\2y-3&10&1&-10&-1&5&2&-5&-2\\x&0&9&-2&-11&1&4& -3&-6\\y&\text{loại}&2&\text{loại}&1&4&\text{loại}&-8&\text{loại}\end{array}\right]$

$\text{c)xy } – 3\text x = -19 $

$\text{x(y – 3)} = -19 $

$-19=1.(-19)=(-1).19$

$\left[\begin{array}{ccc}x&1&-19&-1&19\\y-3&-19&1&19&-1\\y&-21&-2&16&-4\end{array}\right]$

$d)3x+4y-xy=16$

$⇔3x-xy+4y-12=16-12$

$⇔(3x-xy)+(4y-12)=4$

$⇔x(3-y)+4(y-3)=4$

$⇔x(3-y)-4(3-y)=4$

$⇔(x-4)(3-y)=4$

$4=1.4=2.2=(-2).(-2)=(-1).(-4)$

$\left[\begin{array}{ccc}x-4&1&4&2&-2&-1&-4\\3-y&4&1&2&-2&-4&-1\\x&5&8&6&2&3&-1\\y&-1&2&1&5&7&4\end{array}\right]$