1/ $(x+1)(y-2)=1$
$→(x+1);(y-2)\in Ư(1)=\{±1\}$
$→$ Ta có bảng:
$\begin{array}{|c|c|c|}\hline x+1&1&-1\\\hline y-2&1&-1\\\hline x&0&-2\\\hline y&3&1\\\hline\end{array}$
Vậy $(x;y)=\{(0;3);(-2;1)\}$
2/ $(x-5)(y-1)=2$
$→(x-5);(y-1)\in Ư(2)=\{±1;±2\}$
$→$ Ta có bảng:
$\begin{array}{|c|c|c|}\hline x-5&1&-1&2&-2\\\hline y-1&2&-2&1&-1\\\hline x&6&4&7&3\\\hline y&3&-1&2&0\\\hline\end{array}$
Vậy $(x;y)=\{(6;3);(4;-1);(7;2);(3;0)\}$
3/ $(x+4)(y-2)=3$
$→(x+4);(y-2)\in Ư(3)=\{±1;±3\}$
$→$ Ta có bảng:
$\begin{array}{|c|c|c|}\hline x+4&1&-1&3&-3\\\hline y-2&3&-3&1&-1\\\hline x&-3&-5&-1&-7\\\hline y&5&-1&3&1\\\hline\end{array}$
Vậy $(x;y)=\{(-3;5);(-5;-1);(-1;3);(-7;1)\}$
4/ $(x-4)(y+3)=-4$
$→(x-4);(y+3)\in Ư(-4)=\{±1;±2;±4\}$
$→$ Ta có bảng:
$\begin{array}{|c|c|c|}\hline x-4&1&-1&2&-2&4&-4\\\hline y+3&-4&4&-2&2&-1&1\\\hline x&5&3&6&2&8&0\\\hline y&-7&1&-5&-1&-4&-2\\\hline\end{array}$
Vậy $(x;y)=\{(5;-7);(3;1);(6;-5);(2;-1);(8;-4);(0;-2)\}$
5/ $(x+3)(y-6)=-5$
$→(x+3);(y-6)\in Ư(-5)=\{±1;±5\}$
$→$ Ta có bảng:
$\begin{array}{|c|c|c|}\hline x+3&1&-1&5&-5\\\hline y-6&-5&5&-1&1\\\hline x&-2&-4&2&-8\\\hline y&1&11&5&7\\\hline\end{array}$
Vậy $(x;y)=\{(-2;1);(-4;11);(2;5);(-8;7)\}$
