Question

va roooooooog e super urgent

Answer 1

Explicație pas cu pas:

a)

x-2≠0 => x ≠ 2

$\dfrac{2x + 5}{x - 2} = \dfrac{2x - 4 + 9}{x - 2} = \dfrac{2(x - 2) + 9}{x - 2} = \\ = 2 + \dfrac{9}{x - 2} \in \mathbb{Z} \implies \dfrac{9}{x - 2} \in \mathbb{Z}$

$(x - 2) \in \mathcal{D}_{9} \iff (x - 2) \in \Big\{ - 9; - 3; - 1; 1; 3; 9\Big\}\ \ \Big|+2$

$\implies x \in \Big\{ - 7; - 1; 1; 3; 5; 11\Big\}$

b)

5x-1≠0

$\dfrac{4x + 3}{5x - 1} \in \mathbb{Z}$

dacă: (5x-1) îl divide pe (4x+3), atunci: (5x-1) îl divide și pe: 5×(4x+3)

$\dfrac{5(4x + 3)}{5x - 1} = \dfrac{20x + 15}{5x - 1} = \dfrac{4(5x - 1) + 19}{5x - 1} = \\ = 4 + \dfrac{19}{5x - 1} \in \mathbb{Z} \implies \dfrac{19}{5x - 1} \in \mathbb{Z}$

$(5x - 1) \in \mathcal{D}_{19} \iff (5x - 1) \in \Big\{ - 19; - 1; 1; 19\Big\}\ \ \Big|+1$

$5x \in \Big\{ - 18; 0; 2; 20\Big\}$

$x \in \mathbb{Z} \implies 5x \in \mathbb{Z}$

$5x \in \Big\{0; 20\Big\} \implies x \in \Big\{0; 4\Big\}$