[tex]\displaystyle \Big| \frac{n}{2n+1}- \frac{1}{2} \Big| \ \textless \ \frac{1}{100} \Leftrightarrow \\ \\ \Leftrightarrow - \frac{1}{100}\ \textless \ \frac{n}{2n+1}- \frac{1}{2}\ \textless \ \frac{1}{100} \Leftrightarrow \\ \\ \Leftrightarrow -\frac{1}{100}+ \frac{1}{2} \ \textless \ \frac{n}{2n+1}\ \textless \ \frac{1}{100}+ \frac{1}{2} \Leftrightarrow \\ \\ \Leftrightarrow \frac{49}{100}\ \textless \ \frac{n}{2n+1}\ \textless \ \frac{51}{100}. \\ \\ Deoarece~n \in N \Rightarrow 2n+1\ \textgreater \ 0 \Rightarrow 49(2n+1)\ \textless \ 100n\ \textless \ 51(2n+1)} .[/tex]
[tex]\displaystyle i)~49(2n+1)\ \textless \ 100n \Leftrightarrow 98n+49\ \textless \ 100n \Leftrightarrow 49\ \textless \ 2n \Rightarrow n\ \textgreater \ \frac{49}{2} . \\ \\ ii)~100n\ \textless \ 51(2n+1) \Leftrightarrow 100n\ \textless \ 102n+51 \Leftrightarrow 0\ \textless \ 2n+51. \\ \\ Din~i)~am~obtinut~ca~n\ \textgreater \ \frac{49}{2}.~Partea~a~doua~a~inegalitatii~(~ii)~)~se~ \\ \\ verifica~pentru~orice~valoare~naturala~a~lui~n.\\ \\ Solutiile~inecuatiei~sunt~deci~n \in \{ 25,26,27,... \} \rightarrow \boxed{25}~este~cea~mai~ \\ \\ mica.[/tex]
