`(2x +1)(y-5)=12`
`⇔(2x+1)(y-5)∈U(12){ĐK:x,y∈N^*}`
`⇔12=1.12=2.6=3.4`
`⇔2x +1` là số lẻ nên ⇔ \(\left[ \begin{array}{l}2x +1=1\\2x +1=3\end{array} \right.\)
Ta có:
⇔\(\left[ \begin{array}{l}2x +1=1\\2x +1=3\end{array} \right.\) ⇔\(\left[ \begin{array}{l}y-5=12\\y-5=4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=17\\x=9\end{array} \right.\)
