Question

cum se rezolva 009? ajutoor

Answer 1

$[\frac{1}{x}]\in \mathbb{Z} \Rightarrow \frac{1}{[x]} \in \mathbb{Z}, \ daca \ si \ numai \ daca \ 1 \ este \ divizibil \ cu \ [x]\\ \\ \Rightarrow [x]\in \{-1; \ 1\}\\ \\ Cazul \ 1:\\ \\ partea \ intreaga \ a \ lui \ x \ este \ -1 \Rightarrow x\in [-1; 0)\\ \\ \Rightarrow \frac{1}{x}\in (-oo; \ -1],\ iar\ [\frac{1}{x}]=\frac{1}{[x]} \ doar \ daca \ x=-1\\ \\ Cazul \ 2:\\ \\ partea\ intreaga \ a \ lui \ x \ este \ 1 \Rightarrow x\in (0; \ 1]\\ \\ \Rightarrow \frac{1}{x}\in [1; \ +oo),\ iar \ [\frac{1}{x}]=\frac{1}{[x]} \ doar \ daca \ x=1\\ \\ Deci \ [\frac{1}{x}]=\frac{1}{[x]} \ daca \ x \in \{-1; \ 1\}$

Answer 2

$\Big[\dfrac{1}{x}\Big] = \dfrac{1}{[x]}\\ \\\\ \boxed{1}\quad x \in (-1,1) \Rightarrow \dfrac{1}{[x]} = \dfrac{1}{0}\quad (F) \\\\ \boxed{2}\quad x \in \{-1,1\}\quad (A) \\ \\ \boxed{3}\quad x\in (-\infty, -1)\cup(1, +\infty)\Rightarrow \\ \\ \Rightarrow \Big[\dfrac{1}{ (-\infty, -1)\cup (1,+\infty)}\Big] = \dfrac{1}{\Big[ (-\infty, -1)\cup (1,+\infty)\Big]}$

$\Rightarrow \Bigg[\Big(-0,-0.(9)\Big)\cup\Big(0, 0.(9)\Big)\Bigg] = \{...,-5,-4,-3,-2,1,2,3,...\}\\ \\ \Rightarrow \{-1\} \cup\{0\} = \{...,-5,-4,-3,-2,1,2,3,...\}\quad (F)$

$\\\\\Rightarrow S = \{-1,1\} \Rightarrow\,d)\quad \{-1,1\}\quad (corect)$