Question

Determinati multimea

Answer 1

Răspuns:

A={-3;-1;0;2;3;5}

Answer 2

$\frac{x + 3}{x -1} \in\mathbb{Z } \\ \\ 1 + \frac{4}{x-1} \in\mathbb{Z} \\ \\ \implies \: 1 + \frac{4}{x-1 } \in\mathcal{D}_{4} \\ \\ \implies \: 1 + \frac{4}{x -1} \in\{\pm1;\pm2;\pm4\} \\ \\ \frac{4}{x -1} \in\{ \pm1;\pm2;\pm4\}\\ \\ x-1\in\{\frac{4}{-1}; \frac{4}{1}; \frac{4}{-2}; \frac{4}{2}; \frac{4}{-4}; \frac{4}{4}\}\\ \\ x-1\in \{-4;4;-2;2;-1;1\}\\ \\ \boxed{x\in\{ -3;5;-1;3;0;2\}}$