Đáp án:
Giải thích các bước giải:
`a)`
`(x-3).(2y+1)=7`
`=> x-3; 2y+1 in Ư(7)`
Ta có bảng :
$\begin{array}{|c|c|}\hline x-3&-7&-1&1&7\\\hline 2y+1&-1&-7&7&1\\\hline x&-4&2&4&10\\\hline y&-1&-4&3&0\\\hline\end{array}$
Vậy `(x;y) in { (-4;-1); (2;-4); (4;3); (10;0) }`
`b)`
`(2x+1).(3y-2)=-55`
`=> 2x+1; 3y-2 in Ư(-55)`
Ta có bảng :
$\begin{array}{|c|c|}\hline 2x+1&-55&-11&-5&-1&1&5&11&55\\\hline 3y-2&-1&-5&-11&-55&55&11&5&1\\\hline x&-28&-6&-3&-1&0&2&5&27\\\hline y&1&\dfrac{7}{3}_\text{ktm}&\dfrac{13}{3}_\text{ktm}&19&-\dfrac{53}{3}_\text{ktm}&-3&-1&\dfrac{1}{2}_\text{ktm} \\\hline\end{array}$
Vậy `(x;y) in { (-28;1);(-1;19);(2;-3);(5;-1)}`
