[tex]\it \dfrac{[x+1]}{3}=x-\dfrac{3}{2}\Big|_{\cdot6} \Rightarrow2[x+1]=6x-9 \Rightarrow [x+1]=\dfrac{6x-9}{2}\\ \\ \\ Not\breve am\ \dfrac{6x-9}{2}=k\in\mathbb{Z} \Rightarrow x=\dfrac{2k+9}{6}\ \ \ \ \ (1)[/tex]
Ecuația devine:
[tex]\it\Big[\dfrac{2k+9}{6}+1\Big]=k \Rightarrow \Big[\dfrac{2k+15}{6}\Big]=k \Rightarrow k\leq\dfrac{2k+15}{6}<k+1\Big|_{\cdot6} \Rightarrow \\ \\ \\ \Rightarrow 6k\leq2k+15<6k+6|_{-15} \Rightarrow 6k-15\leq2k<6k-9|_{-6k} \Rightarrow \\ \\ \\ \Rightarrow -15\leq-4k<-9|_{:(-4)} \Rightarrow \dfrac{9}{4}<k\leq\dfrac{15}{4} \Rightarrow 2,25<k<3,75 \Rightarrow k=3\ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow x=\dfrac{6+9}{6}=\dfrac{\ 15^{(3}}{6}=\dfrac{5}{2}\\ \\ \\ S_{r(x)}=\Big\{\dfrac{5}{2}\Big\}[/tex]
