Question

Va rog urgent!!!! Mulțumesc anticipat!

Answer 1

Răspuns:

$x \in \Big\{\frac{2(k+1)}{1-k}\Big | k \in \mathbb{Z} \setminus \{1\}\Big\}$

Explicație pas cu pas:

$\frac{x-2}{x+2}\in \mathbb{Z}.\\\textrm{Fie k un numar egal cu valoarea raportului.}\\\frac{x-2}{x+2} = k, k\in \mathbb{Z}\\x-2 = k(x+2)\\x-2 = kx + 2k\\x-kx = 2k + 2\\x(1-k) = 2(k+1)\\x = \frac{2(k+1)}{1-k}, 1-k \neq 0 \iff k \neq 1\Rightarrow \boxed{x \in \Big\{\frac{2(k+1)}{1-k}\Big | k \in \mathbb{Z} \setminus \{1\}\Big\}}$