Question

Determinaţi valorile întregi ale lui x pentru care x-3 supra x+2 este număr întreg.
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Answer 1

[tex] \frac{x-3}{x+2} = \frac{x+2-2-3}{x+2} = \frac{x+2}{x+2} - \frac{5}{x+2} =1-\frac{5}{x+2} =>\frac{5}{x+2} \in Z\\ x+2\in D_5=>x+2\in\{\pm5,\pm1\}\\ x+2=5=>x=3\\ x+2=-5=>x=-7\\ x+2=1=>x=-1\\ x+2=-1=>x=-3\\ x\in\{-7,-3,-1,3\}[/tex]

Answer 2

$\frac{x-3}{x+2} \in \mathbb{Z} \\\\ x \in \mathbb{Z} \\\\ \Longrightarrow x+2|x-3 \\\\ x+2|x+2 \\\\ x+2|x+2-x+3 \\\\ x+2|5 \\ x \in \mathbb{Z} \ \ \Rightarrow x+2 \in \{\pm 1; \pm5 \} \\\\\\ 1) x+2=1 \\\\ x=-1 \\\\\\ 2) x+2=-1 \\\\ x=-3 \\\\\\ 3) x+2=5 \\\\ x=3 \\\\\\ 4)x+2=-5 \\\\ x=-7 \\\\\\ \boxed{ x=\{-7;-3;-1;3 \}}$