[tex]1)$ $ |x^2-1|\cdot (7-|2x+1|)\ \textgreater \ 0 \\ \\ $ \ $ |x^2-1| \ \textgreater \ 0,\quad \forall x \in \mathbb_{Z}^* $ \\ $ $ deci, problema noastra ramane ca $ 7-|2x+1| \ \textgreater \ 0 \\ \\ \Rightarrow -|2x+1| \ \textgreater \ -7 \Rightarrow |2x+1| \ \textless \ 7 \Rightarrow -7 \ \textless \ 2x+1\ \textless \ 7\Big{|}-1 \Rightarrow \\ \Rightarrow -8\ \textless \ 2x\ \textless \ 6\Big{|}:2 \Rightarrow -4\ \textless \ x\ \textless \ 3 \Rightarrow x \in \{-3,-2,-1,1,2\} \\ \\ \Rightarrow $ card(A) = \{5\}[/tex]
[tex]($fara zero, deoarece x \in \mathbb_{Z}^*, $ steluta inseamna ca ia toate valorile din $ \mathbb_{Z} $ \\ inafara de 0)[/tex]
[tex]2) $ $ \sqrt{n^2-6n+34}\in \mathbb_{N} $ \Rightarrow \sqrt{n^2-6n+9+25}\in \mathbb_{N} $ \\ \\ \Rightarrow \sqrt{(n-3)^2+25}\in \mathbb_{N} \Rightarrow $ \ $ n=3[/tex]
