[tex]\Big[ \frac{2n^2}{n+1} \Big]=n \in Z. \\ \\ n \neq -1. \\ \\ Avem:~n \leq \frac{2n^2}{n+1}\ \textless \ n+1. \\ \\ Cazul~1:~n+1 \ \textgreater \ 0 \Leftrightarrow n\ \textgreater \ -1. \\ \\ Rezulta~n^2+n \leq 2n^2\ \textless \ n^2+2n+1 \Leftrightarrow n \leq n^2\ \textless \ 2n+1. \\ \\ De~aici~avem~n(1-n) \leq 0~si~n^2-2n-1\ \textless \ 0 \Leftrightarrow \\ \\ \Leftrightarrow n^2-2n+1\ \textless \ 2 \Leftrightarrow (n-1)^2\ \textless \ 2 \Rightarrow - \sqrt{2}\ \textless \ n-1\ \textless \ \sqrt{2}. [/tex]
[tex]Iar~cum~n \in Z \Rightarrow n \in \{0;1;2 \}.~Si~ramane~sa~le~verificam, \\ \\ inlocuindu-le~in~ecuatia~initiala.~Toate~aceste~valori~ \\ \\ verifica~relatia,~deci~sunt~soluii.[/tex]
[tex]Cazul~2:~n+1\ \textless \ 0 \Leftrightarrow n\ \textless \ -1. \\ \\ In~acest~caz,~din~n \leq \frac{2n^2}{n+1}\ \textless \ n+1~va~rezulta: \\ \\ n^2+n \geq 2n^2\ \textgreater \ n^2+2n+1 \Leftrightarrow n \geq n^2\ \textgreater \ 2n+1. \\ \\ n \geq n^2 \Leftrightarrow n(1-n) \geq 0 ,~fals!~deoarece~n\ \textless \ 0~si~1-n\ \textgreater \ 0. \\ \\ Deci~acest~caz~nu~prezinta~solutii. \\ \\ Prin~urmare~solutiile~ecuatiei~sunt~0,1,2.[/tex]
