Question

Vă rog exercițiul 6....

Answer 1

2x+1 divide pe 7x-5
2x+1 divide pe14x-10
(14x-10)/(2x+1)∈Z
(14x+7-17)/(2x+1)∈z

(14x+7/(2x+1)-17/(2x+1)∈Z
2-17/(2x+1)∈Z
 -17/(2x+1)∈Z
17/(2x+1)

2x+1∈D17Z
2x+1=1...x=0
2x+1=-1...x=-1
2x+1=17....x=8
2x+1=-17....x=-9

x∈{-9;-1;0;8}

Answer 2

[tex]\it \dfrac{7x-5}{2x+1} \in \mathbb{Z},\ \ x\in \mathbb{Z}, \ \ \ x=? \\\;\\ \\\;\\ \dfrac{7x-5}{2x+1} \in \mathbb{Z} \Rightarrow 2x+1|7x-5 \Rightarrow 2x+1|2(7x-5) \Rightarrow \\\;\\ \\\;\\ \Rightarrow 2x+1|14x-10 \ \ \ (1)[/tex]


[tex]\it Dar,\ \ 2x+1|2x+1 \Rightarrow 2x+1|7(2x+1) \Rightarrow 2x+1|14x+7 \ \ \ (2) \\\;\\ \\\;\\ (1),\ (2) \Rightarrow 2x+1| 14x+7-(14x-10) \Rightarrow 2x+1|14x+7-14x+10 \\\;\\ \\\;\\ \Rightarrow 2x+1|17 \Rightarrow 2x+1 \in D_{17} \Rightarrow 2x+1 \in\{\pm1,\ \pm17\} \Rightarrow[/tex]


[tex]\it \Rightarrow 2x+1\in \{-17,\ -1,\ 1,\ 17\}|_{-1} \Rightarrow 2x\in \{-18,\ -2,\ 0,\ 16\}|_{:2} \Rightarrow \\\;\\ \\\;\\ \Rightarrow x\in\-\{-9,\ -2,\ 0,\ 8\}.[/tex]