[tex][x+1]=2x-1\\ \\ $Stim ca [u] = k \Rightarrow k\leq u\ \textless \ k+1, \quad k\in \mathbb_{Z}$ $ \\ \\ $Noi avem: \quad $ [x+1] = 2x-1 \\ \\ \Rightarrow \\ 2x-1\leq x+1\ \textless \ (2x-1)+1 \\ \\ 2x-1\leq x+1\ \textless \ 2x \Big|-2x \\ \\ -1 \leq -x+1\ \textless \ 0 \Big|-1 \\ \\ -2 \leq -x\ \textless \ -1 \Big|\cdot (-1) \\ \\ 2\geq x\ \textgreater \ 1 \\ \\ $Dar, $ 2x-1 = k \Rightarrow 2x = k+1 \Rightarrow x = \dfrac{k+1}{2}, \quad k\in \mathbb_{Z}$ $\\ \\ $Inlocuim: \\ \\ $ 2\geq \dfrac{k+1}{2}\ \textgreater \ 1 \Big|\cdot 2 \\ \\ 4\geq k+1\ \textgreater \ 2 \Big|-1 \\ \\ 3\geq k\ \textgreater \ 1 \\ \\ 1\ \textless \ k\leq3[/tex]
[tex]\Rightarrow k \in \Big\{2,3\Big\} \\ \\ \bullet $ $ k=2\Rightarrow x = \dfrac{2+1}{2} \Rightarrow \boxed{x = \dfrac{3}{2}} \\ \\ \bullet $ $ k = 3 \Rightarrow x = \dfrac{3+1}{2} \Rightarrow \boxed{x = 2}[/tex]
[tex]\Rightarrow \boxed{S = \Big\{\dfrac{3}{2},2\Big\}}[/tex]
[tex]\{x\}=\dfrac{1}{3} \\ \\ $ Din formula fundamentala: \quad $ x = [x]+\{x\}\Rightarrow \\ \\ \Rightarrow \{x\} = x-[x] \\ \\ \Rightarrow x-[x] = \dfrac{1}{3} \Rightarrow [x] = x-\dfrac{1}{3}\\ \\ $Stim ca: \quad $[u] = k \Rightarrow k\leq u\ \textless \ k+1 \\ \\ \Rightarrow x-\dfrac{1}{3}\leq x\ \textless \ \Big(x-\dfrac{1}{3}\Big)+1 \Big|\cdot 3 \\ \\ 3x-1\leq 3x\ \textless \ 3x-1+3\Big|-3x \\ \\ -1\leq 0\cdot x\ \textless \ 2 \Rightarrow x\in \mathbb_{R},$ Dar, $ x-\dfrac{1}{3}\in \mathbb_{Z}$ \Rightarrow \\ \\ $ [/tex]
[tex]\Rightarrow x-\dfrac{1}{3} = k,\quad k\in \mathbb_{Z} $ $ \\ \\ \Rightarrow x = k+\dfrac{1}{3} \Rightarrow \boxed{x = \dfrac{3k+1}{3},\quad k\in \mathbb_{Z}}[/tex]
