Ta có :
\(2x+15⋮x+1\)
Mà \(x+1⋮x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+15⋮x+1\\2x+2⋮x+1\end{matrix}\right.\)
\(\Leftrightarrow13⋮x+1\)
Vì \(x\in Z\Leftrightarrow x+1\in Z;x+1\inƯ\left(13\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=13\\x+1=-13\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=12\\x=-14\end{matrix}\right.\)
Vậy ...
Ta có : 2x + 15 \(⋮\) x + 1 . Mà x + 1\(⋮\) x + 1
\(\Rightarrow\) 2x + 15 \(⋮\) x + 1
2x + 2 \(⋮\) x + 1
\(\Rightarrow\) 13 \(⋮\) x + 1
Vì x \(\in\) Z nên x + 1 \(\in\) Z ; x \(\in\) Ư(13)
\(\Rightarrow\) x + 1 = \(\pm\) 1 \(\Leftrightarrow\) x = 0 ; -2
\(\Rightarrow\) x + 1 = \(\pm\) 13 \(\Leftrightarrow\) x = 12 ; -14