Ta có 2x + 7 = 2 ( x + 1 ) +5
Vì 2 ( x + 1 ) chia hết cho x + 1
=> 5 chia hêt x + 1
hay x + 1 thuộc Ư(5) = { 1 ; 5 }
=> x thuộc { 0 ; 4 }
\(2x+7⋮x+1\)
\(\Rightarrow2x+2+5⋮x+1\)
\(\Rightarrow2\left(x+1\right)+5⋮x+1\)
mà \(2.\left(x+1\right)⋮x+1\)
\(\Rightarrow5⋮x+1\)
\(\Rightarrow x+1\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(x+1=1\Rightarrow x=0\)
\(x+1=-1\Rightarrow x=-2\)
\(x+1=5\Rightarrow x=4\)
\(x+1=-5\Rightarrow x=-6\)
\(\Rightarrow x\in\left\{0;-2;4;-6\right\}\)
\(2x+7⋮x+1\)
\(\Rightarrow2x+2+5⋮x+1\)
\(\Rightarrow2\left(x+1\right)+5⋮x+1\)
\(\Rightarrow2\left(x+1\right)⋮x+1\)
\(\Rightarrow5⋮x+1\)
\(\Rightarrow x+1\inƯ\left(5\right)\)
\(\RightarrowƯ\left(5\right)=\left\{1;5\right\}\)
\(\Rightarrow\orbr{\begin{cases}x+1=1\Rightarrow x=0\\x+1=5\Rightarrow x=4\end{cases}}\)
\(\Rightarrow x\in\left\{0;4\right\}\)
\(\left(2x+7\right)⋮\left(x+1\right)\)
\(\Rightarrow\left(2x+7\right)⋮2\left(x+1\right)\)
\(\Rightarrow\left(2x+1\right)⋮2x+2\)
\(\Rightarrow\left[\left(2x+7\right)-\left(2x+2\right)\right]⋮\left(x+1\right)\)
\(\Rightarrow\left[2x+7-2x-2\right]⋮\left(x+1\right)\)
\(\Rightarrow5⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)\inƯ_{\left(5\right)}=\left\{\pm1;\pm5\right\}\)
Mà \(x\in N\Rightarrow\left(x+1\right)\in N=\left\{1;5\right\}\)
\(\Rightarrow x\in\left\{0;4\right\}\)
