Giải thích các bước giải:
a) `(x-2)(y+3)=15`
`⇒ x-2; y+3 ∈ Ư(15)={±1;±3;±5;±15}`
Ta có bảng sau:
$\begin{array}{|c|c|} \hline x-2&-15&-5&-3&-1&1&3&5&15\\\hline x &-13&-3&-1&1&3&5&7&17\\\hline y+3&-1&-3&-5&-15&15&5&3&1\\\hline y &-4&-6&-8&-18&12&2&0&-2\\\hline\end{array}$
Vậy `....`
b) `(2x - 1)(y-4)=-13`
`⇒ 2x - 1; y - 4 ∈ Ư(-13)={±1;±13}`
Ta có bảng sau:
$\begin{array}{|c|c|}\hline 2x -1&-13&-1&1&13\\\hline x&-6&0&1&7\\\hline y-4&1&13&-13&-1\\\hline y &5&17&-9&3\\\hline \end{array}$
Vậy `.......`
c) `(3x + 2)(1-y)=-7`
`⇒ 3x + 2; 1 - y ∈ Ư(-7)={±1;±7}`
Ta có bảng sau:
$\begin{array}{|c|c|}\hline 3x+2&-7&-1&1&7\\\hline x&-3&-1&/&/\\\hline 1-y&1&7&-7&-1\\\hline y&0&-6&8&2\\\hline\end{array}$
Vậy `(x;y) = (-3;0);(-1;-6)`
d) $5xy-5x+y=5\\⇔5x(y-1)+y-1=4\\⇔ 5x(y-1)+(y-1)=4\\⇔(5x+1)(y-1)=4\\\Rightarrow 5x+1;y-1 \in Ư(4)=\{\pm 1;\pm 2; \pm 4\}$
Ta có bảng sau:
$\begin{array}{|c|c|}\hline 5x+1&-4&-2&-1&1&2&4\\\hline x&-1&/&/&0&/&/\\\hline y-1&-1&-2&-4&4&2&1\\\hline y&0&-1&-3&5&3&2\\\hline \end{array}$
Vậy `(x;y) = (-1;0);(0;5)`
