Question

Buna! Ma puteti ajuta va rogggg? ​

Answer 1

Răspuns:

Explicație pas cu pas:

a)

$E(x) = \bigg(\dfrac{x+5}{x+4} - \dfrac{x-7}{x^{2}+3x-4} - \bigg) : \bigg(1+\dfrac{3}{x-1} \bigg)$

$= \bigg(\dfrac{^{x-1)} x+5}{x+4} - \dfrac{x-7}{(x-1)(x+4)}\bigg) : \dfrac{x-1+3}{x-1}$

$= \dfrac{(x-1)(x+5)-(x-7)}{(x-1)(x+4)} : \dfrac{x+2}{x-1}$

$= \dfrac{x^{2}+5x-x-5-x+7}{(x-1)(x+4)} \cdot \dfrac{x-1}{x+2}$

$= \dfrac{x^{2}+3x+2}{x+4} \cdot \dfrac{1}{x+2} = \dfrac{(x+2)(x+1)}{x+4} \cdot \dfrac{1}{x+2}$

$= \dfrac{x+1}{x+4} \cdot \dfrac{1}{1} = \dfrac{x+1}{x+4}$

b)

$E(x) = \dfrac{x+1}{x+4} = \dfrac{x+4-3}{x+4} = 1 -\dfrac{3}{x+4} \in \mathbb{Z^{*}}$

$\implies \dfrac{3}{x+4} \in \mathbb{Z^{*}} \iff (x+4) \in \mathcal{D}_{3}$

$(x+4) \in \{-3;-1;1;3\} \ \Big|(-4)$

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

$E(x) \neq 0 \implies x \in \{-7;-5;-3\}$