Tham khảo
`c) \frac{-3+x}{4}=\frac{-3}{y-2}`
`⇒\frac{x-3}{4}=\frac{-3}{y-2}`
`⇒(x-3)(y-2)=4.-3`
`⇒(x-3)(y-2)=-12`
Ta có bảng:
$\left[\begin{array}{ccc}x-3&1&-1&2&-2&3&-3&4&-4&6&-6&12&-12\\x&4&2&5&1&6&0&7&-1&9&-3&15&-9\\y-2&12&-12&6&-6&4&-4&3&-3&2&-2&1&-1\\y&14&-10&8&-4&6&-2&5&-1&4&0&3&1\end{array}\right]$
Vậy `y∈ {14,-10,8,-4,6,-2,5,-1,4,0,3,1}`
`\text{©CBT}`