[tex]\\ $Avem numarul $3+\dfrac{1}{\sqrt3-3},\quad $pornim de la inecuatia $ 1\ \textless \ \sqrt3\ \textless \ 2. \\ \\ \\ 1\ \textless \ \sqrt3\ \textless \ 2 \Big|-3 \\ \\ -2\ \textless \ \sqrt3-3\ \textless \ -1\Big|^{-1} \\ \\ \dfrac{1}{-2}\ \textgreater \ \dfrac{1}{\sqrt3-3}\ \textgreater \ \dfrac{1}{-1} \\ \\ -\dfrac{1}{2}\ \textgreater \ \dfrac{1}{\sqrt3-3}\ \textgreater \ -1\Big|+3 \\ \\ -\dfrac{1}{2}+3\ \textgreater \ \dfrac{1}{\sqrt3-3}+3\ \textgreater \ 2 \\ \\[/tex]
[tex]2\ \textless \ \dfrac{1}{\sqrt3-3}+3\ \textless \ -\dfrac{1}{2}+3 \\ \\ 2\ \textless \ 3+\dfrac{1}{\sqrt3-3}\ \textless \ 2,5\quad \quad $(ne intereseaza doar de 2) \\ \\ \Rightarrow $ cel mai mare numar intreg mai mic decat $ 3+\dfrac{1}{\sqrt3-3} $ este 2.[/tex]
