Đặt \(A=\dfrac{15-2n}{n+1}=-2+\dfrac{17}{n+1}\)
Muốn \(\left(15-2n\right)⋮\left(n+1\right)\) thì \(A\in Z\)
\(=>\dfrac{17}{n+1}\in Z\)
\(=>n+1\inƯ\left(17\right)\)
\(\cdotƯ\left(17\right)=\left\{1;-1;17;-17\right\}\)
\(=>n+1=1;n+1=-1;n+1=17;n+1=-17\)
\(=>n=\left\{0;-2;16;-18\right\}\)
Mà \(n\le7\)
\(=>n\in\left\{0;-2\right\}\)