ta có: \(3x+17⋮x+1\Rightarrow3x+3+14⋮x+1\)
mà 3x+3 chia hết cho x+1 \(\Rightarrow14⋮x+1\Rightarrow x+1\in\left\{1;2;7;14\right\}\Rightarrow x\in\left\{0;1;6;13\right\}\)
\(17+3x⋮x+1\)
\(\Leftrightarrow3x+17⋮x+1\)
Mà \(x+1⋮x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+17⋮x+1\\3x+3⋮x+1\end{matrix}\right.\)
\(\Leftrightarrow14⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(14\right)\)
Mà \(x\in N\Leftrightarrow x+1\in N\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=14\\x+1=1\\x+1=2\\x+1=7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=13\\x=0\\x=1\\x=6\end{matrix}\right.\)
Vậy ..
