Question

determina numerele întregi x pentru care numărul
$\frac{3x + 5}{2x - 1}$
este număr intreg​

Answer 1

Explicație pas cu pas:

x € Z pentru care[ (3x +5) / (2x -1) ] € Z

(3x +5) / (2x -1) € Z < = > (3x +5) / (2x -1) = ×/- 1 =>

=> { (3x+5)/ (2x -1) = 1 =>

{ (3x+5)/(2x-1)= - 1

= > { 3x +5 =2x -1 => { x = - 6 € Z

{ 3x +5 = -2x +1 { x = -4 / 5 nu € Z.

( 3x +5)/ (2x -1) = (2x - 1)/ (2x - 1) + ( x +6) / (2x- 1) =

= [1 + (x +6)/( 2x - 1) ] € Z < =>

< = > (x+6)/ ( 2x - 1) €Z => ( x+6) / (2x - 1) = +/- 1

=> { (x +6)/ (2x - 1) = 1 => { x +6 = 2x- 1 =>

{ (x+6) / (2x - 1) = -1 { x +6 = -2x +1

=> { x = 7 € Z

{ 3x = -5 /3 nu € Z

Deci ; x € { -6 , 7 }

Answer 2

...$\it \dfrac{3x+5}{2x-1}\in\mathbb{Z} \Rightarrow 2x-1\Big |3x+5 \Rightarrow 2x-1\Big|(3x+5)\cdot2 \Rightarrow 2x-1\Big| 6x+10\ \ \ \ \ \ (1)\\ \\ \\ Dar,\ 2x-1\Big | 2x-1|_{\cdot3} \Rightarrow 2x-1\Big |6x-3\ \ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow 2x-1\Big |6x+10-6x+3 \Rightarrow 2x-1\Big |13 \Rightarrow 2x-1\in D_{13} \Rightarrow \\ \\ \\ \Rightarrow 2x-1\in\{-13,\ -1,\ 1,\ 13\}|_{+1} \Rightarrow 2x\in\{-12,\ 0,\ 2,\ 14 \}|_{:2} \Rightarrow \\ \\ \\ \Rightarrow x\in\{-6,\ 0,\ 1,\ 7\}$