Question

Umpic de ajutor ! Simplu ! Daca [x+1]=2x-1 in ce multime/interval apartine x ?
Acelasi lucru la {x}=1/3

Answer 1

[x+1]=[x]+1

[x]+1=2x-1
[x]=2x-2

[x]∈Z, -2∈Z⇒2x∈Z (1)

rezolvand grafic **** am obtinut solutiile x=3/2 si x=2
verificare
[3/2]=2*3/2-2
1=3-2 adevarat

sau asa cum sunt date



[2]=2*2-2
2=4-2 , adevarat

am intersectat graficul functie [x] cu graficul functiei 2x-2

analitic, sorry,momentan nu am inspiratie


b)la b ) x ∈{n+1/3| n ∈N}∪  {-n-2/3| n ∈N}



a) analitic
 avem
[x]≤x<[x]+1  si
[x]=2x-2

2x-2≤x<2x-2+1 si 2x∈Z

2x-2≤x<2x-1

2x-2-x≤0
x-2≤0
x≤2

x<2x-1
x-2x<-1
-x<-1
x>1
deci x∈(1;2] si 2x∈Z  vezi conditia (1) de sus
singurele solutii sunt x=3/2 si x=2
 deci ACELEASI solutii cu cele obtinute  grafic
YESSSS!

Answer 2

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

$\Rightarrow k \in \Big\{2,3\Big\} \\ \\ \bullet k=2\Rightarrow x = \dfrac{2+1}{2} \Rightarrow \boxed{x = \dfrac{3}{2}} \\ \\ \bullet k = 3 \Rightarrow x = \dfrac{3+1}{2} \Rightarrow \boxed{x = 2}$

$\Rightarrow \boxed{S = \Big\{\dfrac{3}{2},2\Big\}}$



$\{x\}=\dfrac{1}{3} \\ \\ Din formula fundamentala: \quad x = [x]+\{x\}\Rightarrow \\ \\ \Rightarrow \{x\} = x-[x] \\ \\ \Rightarrow x-[x] = \dfrac{1}{3} \Rightarrow [x] = x-\dfrac{1}{3}\\ \\ Stim ca: \quad [u] = k \Rightarrow k\leq u\ \textless \ k+1 \\ \\ \Rightarrow x-\dfrac{1}{3}\leq x\ \textless \ \Big(x-\dfrac{1}{3}\Big)+1 \Big|\cdot 3 \\ \\ 3x-1\leq 3x\ \textless \ 3x-1+3\Big|-3x \\ \\ -1\leq 0\cdot x\ \textless \ 2 \Rightarrow x\in \mathbb_{R}, Dar, x-\dfrac{1}{3}\in \mathbb_{Z} \Rightarrow$

$\Rightarrow x-\dfrac{1}{3} = k,\quad k\in \mathbb_{Z} \\ \\ \Rightarrow x = k+\dfrac{1}{3} \Rightarrow \boxed{x = \dfrac{3k+1}{3},\quad k\in \mathbb_{Z}}$