Ta có:
\(\dfrac{x+13}{x+1}=\dfrac{13\left(x+1\right)}{x+1}=\dfrac{13}{x+1}.\dfrac{x+1}{x+1}=\dfrac{13}{x+1}.1=\dfrac{13}{x+1}\)
Để \(13⋮x+1\)
Hay \(x+1\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
Ta có bảng sau:
\(x+1\) | \(-13\) | \(-1\) | \(1\) | \(13\) |
\(x\) | \(-14\) | \(-2\) | \(0\) | \(12\) |
Vậy \(x\in\left\{-14;-2;0;12\right\}\) thì \(x+13⋮x+1\)
\(#WendyDang\)