Ta co : \(\dfrac{2k+2}{k-3}=\dfrac{2k-6+8}{k-3}=\dfrac{2.\left(k-3\right)+8}{k-3}=\dfrac{2.\left(k-3\right)}{k-3}+\dfrac{8}{k-3}=2+\dfrac{8}{k-3}\)
Để \(\dfrac{2k+2}{k-3}\) nguyên thì \(\dfrac{8}{k-3}\) cũng phải nguyên
\(=>k-3\inƯ\left(8\right)\)
\(=>k-3\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
\(=>k\in\left\{-5;-1;1;2;4;5;7;11\right\}\)
tick cho mk nha
chưa hiểu chỗ nào thì hỏi mk nha