[tex]\it a)\ [x]+[x+1]+[x+2]=24 \Rightarrow [x]+[x]+1 +[x]+2=24 \Rightarrow \\ \\ \Rightarrow 3x+3=24|_{:3} \Rightarrow [x]+1=8|_{-1} \Rightarrow [x]=7 \Rightarrow x\in[7,\ \ 8)\\ \\ \\ b)\ [x+1]=2x-1\ \ \ \ \ (1)\\ \\ (1) \Rightarrow 2x-1=k\in\mathbb{Z} \Rightarrow x=\dfrac{k+1}{2}\ \ \ \ \ \ \ (2)[/tex]
Acum, ecuația devine:
[tex]\it \Big[\dfrac{k+3}{2}\Big]=k \Rightarrow k\leq\dfrac{k+3}{2}<k+1|_{\cdot2} \Rightarrow2k\leq k+3<2k+2|_{-3} \Rightarrow \\ \\ \\ \Rightarrow 2k-3\leq k<2k-1|_{-2k} \Rightarrow -3\leq -k<-1|_{\cdot(-1)} \Rightarrow 1< k\leq3|_{+1} \Rightarrow \\ \\ \\ \Rightarrow 2<k+1\leq4|_{:2} \Rightarrow 1<\dfrac{k+1}{2}\leq2\ \ \ \ \ (3)[/tex]
[tex]\it (2),\ (3) \Rightarrow 1<x\leq2 \Rightarrow x\in(1,\ 2][/tex]
[tex]\it c )\ \ \{2x\}=0 \Rightarrow 2x=k\in\mathbb{Z} \Rightarrow x=\dfrac{k}{2},\ \forall\ k\in\mathbb{Z}[/tex]
