Tham khảo
`\frac{x-3}{2}=\frac{3}{y}(y\ne 0)`
`⇒(x-3)y=6`
`⇒x-3,y∈Ư(6)={±1,±2,±3,±6}`
Ta có bảng:
$\left[\begin{array}{ccc}x-3&1&-1&2&-2&3&-3&6&-6\\x&4&2&5&1&6&0&9&-3\\y&6&-6&3&-3&2&-2&1&-1\end{array}\right]$
Vậy `(x,y)=(4,6);(2,-6);(5,3);(1,-3);(6,2);(0,-2);(9,1);(-3,-1)`
`\text{©CBT}`