Răspuns :
[tex]\dfrac{2x+4}{x+1}\in \mathbb_{Z} $ $\\ \\ \\ \dfrac{2x+4}{x+1} = \dfrac{2x+2+2}{x+1} = \dfrac{2(x+1)+2}{x+1} = \dfrac{2(x+1)}{x+1}+\dfrac{2}{x+1} = \\ \\ = \underset{\in \mathbb_{Z}}{2}$ +\dfrac{2}{x+1} \\ \\ \bullet $ $ \dfrac{2}{x+1}\in \mathbb_{Z}$ $ \Rightarrow (x+1) $ $| $ $2 \Rightarrow (x+1)\in D_2 \Rightarrow \\ \Rightarrow (x+1)\in \Big\{-2,-1,1,2\Big\}\Big|-1 \Rightarrow \\ \Rightarrow x \in \Big\{-2-1,-1-1,1-1,2-1\Big\} \Rightarrow \boxed{x\in \Big\{-3,-2,0,1\Big\}}[/tex]
(2x+4)/(x+1)=(2x+2+2)/(x+1)=
2(x+1)/(x+1)+2/(x+1)=2+2/(x+1)
Pt ca acest numar sa apartina lui Z trebuie ca fractia sa fie numar intreg,Acest lucru se intampla daca x+2 este divizor al lui 2
x+1={-2,-1,1 2}
x+1=-2 x=-3
x+1=-1 x=-2
x+1=1 x=0
x+1=2 x=1
x={-3,-2,0,1}
2(x+1)/(x+1)+2/(x+1)=2+2/(x+1)
Pt ca acest numar sa apartina lui Z trebuie ca fractia sa fie numar intreg,Acest lucru se intampla daca x+2 este divizor al lui 2
x+1={-2,-1,1 2}
x+1=-2 x=-3
x+1=-1 x=-2
x+1=1 x=0
x+1=2 x=1
x={-3,-2,0,1}