Ta có : ( 2.x + 7 ) \(⋮\) ( x + 1)
\(\Leftrightarrow\) 2( x + 1 ) + 5 \(⋮\) ( x + 1 )
\(\Rightarrow\) 5 \(⋮\) ( x + 1 )
\(\Rightarrow\) ( x + 1 ) \(\in\) Ư ( 5 ) = { 1;-1;5;-5 }
* Nếu x + 1 = 1
\(\Rightarrow\) x = 1 - 1 = 0
* Nếu x + 1 = -1
\(\Rightarrow\) x = -1 - 1 = -2
* Nếu x + 1 = 5
\(\Rightarrow\) x = 5 - 1 = 4
* Nếu x + 1 = -5
\(\Rightarrow\) x = -5 - 1 = -6
Vậy x \(\in\) { 0; -2; 4; -6 }
Ta có: \(2x+7⋮x+1\)
\(\Leftrightarrow2x+2+5⋮x+1\)
Mà \(2x+2⋮x+1\)
\(\Rightarrow5⋮x+1\)
\(\Rightarrow x+1\inƯ_{\left(5\right)}\)
\(\Rightarrow x+1\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow\left[\begin{matrix}x+1=-5\\x+1=-1\\x+1=1\\x+1=5\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=-6\\x=-2\\x=0\\x=4\end{matrix}\right.\)
Vậy: \(x=-6;-2;0;4\)
