[tex]\it Vom\ nota\ \dfrac{x-1}{2}=k,\ \ (k\in\mathbb{Z}),\ \Rightarrow x=2k+1\ \ \ \ (1)\\ \\ Ecua\c{\it t}ia\ devine:\\ \\ \Big[\dfrac{3(2x+1)-1}{4}\Big] =k \Rightarrow k\leq\dfrac{6k+2}{4}<k+1|_{\cdot4} \Rightarrow 4k\leq6k+2<4k+4|_{-4k} \Rightarrow\\ \\ \Rightarrow 0 \leq{2k+2} <4|_{-2} \Rightarrow -2\leq2k<2|_{:2} \Rightarrow -1\leq k<1 \Rightarrow k\in\{-1,\ 0\}\ \ \ (2)\\ \\ (1),\ (2) \Rightarrow x\in\{-1,\ \ 1\}[/tex]
