Tìm các số nguyên $x$ và $y$, biết: a) $xy-x+2y-2=5$ b) $xy-2x-3y=3$

`\text{a)}` `xy - x + 2y - 2 = 5`

`<=> (xy + 2y) - (x + 2) = 5`

`<=> y (x + 2) - (x + 2) = 5`

`<=> (x + 2)(y - 1) = 5` 

TH1: \(\left[ \begin{array}{l}x+2=5\\y-1=1\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=3\\y=2\end{array} \right.\) 

TH2: \(\left[ \begin{array}{l}x+2=1\\y-1=5\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-1\\y=6\end{array} \right.\) 

TH3: \(\left[ \begin{array}{l}x+2=-5\\y-1=-1\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-7\\x=0\end{array} \right.\) 

TH4: \(\left[ \begin{array}{l}x+2=-1\\y-1=-5\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-3\\x=-4\end{array} \right.\) 

Vậy `x in {3; -1; -7; -3}` và `y in {2; 6; 0; -4}`

`text{b)}` `xy - 2x - 3y = 3`

`<=> xy - 2x - 3y = - 6 + 9`

`<=> xy - 2x - 3y + 6 = 9`

`<=>  x (y - 2) - 3 (y - 2) = 9`

`<=> (y - 2)(x - 3) = 9`

TH1: \(\left[ \begin{array}{l}y-2=9\\x-3=1\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}y=11\\x=4\end{array} \right.\) 

TH2: \(\left[ \begin{array}{l}y-2=1\\x-3=9\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}y=3\\x=12\end{array} \right.\) 

TH3: \(\left[ \begin{array}{l}y-2=-9\\x-3=-1\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}y=-7\\x=2\end{array} \right.\) 

TH4: \(\left[ \begin{array}{l}y-2=-1\\x-3=-9\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}y=1\\x=-6\end{array} \right.\) 

TH5: \(\left[ \begin{array}{l}y-2=3\\x-3=3\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}y=5\\x=6\end{array} \right.\)

TH6: \(\left[ \begin{array}{l}y-2=-3\\x-3=-3\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}y=-1\\x=0\end{array} \right.\)

Vậy `x in {4; 12; 2; -6; 6; 0}` và `y in {11; 3; -7; 1; 5; -1}`