Ta có
\(2n+7=2n+2+5=2\left(n+1\right)+5\)
Để \(\dfrac{2n+7}{n+1}\) có giá trị nguyên thì
\(2n+7⋮n+1< =>2\left(n+1\right)+5⋮n+1< =>5⋮n+1\)
\(=>\left(n+1\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(n+1=1=>n=0\)
\(n+1=-1=>n=-2\)
\(n+1=5=>n=4\)
\(n+1=-5=>n=-6\)
\(=>n=0;-2;4;-6\)