Vì : \(\frac{x+4}{x+1}\) mà \(\frac{x+1}{x+1}\Rightarrow\frac{\left(x+4\right)-\left(x+1\right)}{x+1}\Rightarrow\frac{x+4-x-1}{x+1}\Rightarrow\frac{3}{x+1}\Rightarrow x+1\inƯ\left(3\right)\)
Mà : \(Ư\left(3\right)=\left\{1;3\right\}\)
+) Nếu x + 1 = 1 => x = 1 - 1 => x = 0
+) Nếu x + 1 = 3 => x = 3 - 1 => x = 2
Vậy \(x\in\left\{0;2\right\}\) thì : \(x+4⋮x+1\)
\(\left(x+4\right)⋮\left(x+1\right)\Leftrightarrow\left(x+1\right)+3⋮\left(x+1\right)\)
Mà \(\left(x+1\right)⋮\left(x+1\right)\) nên để \(\left(x+4\right)⋮\left(x+1\right)\)thì \(3⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)\in\)U(3)
\(\Rightarrow\left(x+1\right)\in\left\{1;3;-1;-3\right\}\)
\(\Rightarrow x\in\left\{0;2;-2;-4\right\}\)
Vậy để \(\left(x+4\right)⋮\left(x+1\right)\)thì \(x\in\left(0;2;-2;-4\right)\)
Ta có : (x+4)=(x+1)+3 chia hết cho( x+1)
mà x+1 chia hết cho x+1=> 3 chia hết(x+1)
=> x+1 ϵ{1;3}
với x+1=1=>x=0
với x+1=3=>x=2
vậy xϵ{0;2}