Question

Va rooooooooog! E URGENT!!! ​

Answer 1

Răspuns:

A={x∈Z║(2x+3)/(x+1)∈Z}

(2x+3)/(x+1)=(2x+2+1)/(x+1)=

(2x+2)/(x+1)+1/(x+1)=2(x+1)/(x+1)+1/(x+1)=

2+1/(x+1)

2 este numar intreg.Pui condditia ca si 1/(x+1) sa fie numar intreg

1/(x+1) este numar intreg daca x+1 este divizor intreg  al lui 1=>

x+1=1=>x=0

si x+1= -1  x= -2

x={-2,0}

Explicație pas cu pas:

Answer 2

$\it\dfrac{2x+3}{x+1} = \dfrac{2x+2+1}{x+1} =\dfrac{2(x+1)}{x+1}+ \dfrac{1}{x+1} =2+ \dfrac{1}{x+1} \in\mathhbb{Z} \Rightarrow \dfrac{1}{x+1}\in\mathbb{Z} \Rightarrow\\ \\ \\ \Rightarrow x+1\in\{-1,\ \ 1\}|_{-1} \Rightarrow x\in\{-2,\ \ 0\}$