Question

cum fac daca e in paranteza patrata ?

Answer 1

 mai SIMPLU  (....)  CLASIC, cum vor profesorii
stiind ca [x]≤x<x+1
avem [(x+1)/5]=(2x-1)/3≤(x+1)/5<(2x-1)/3+1

(2x-1)/3≤(x+1)/5<(2x+2)/3   inmul;tim cu 15
10x-5≤3x+3<10x+10
 
din prima 7x≤8  adica x≤8/7
din a doua7x>-7 adica x>-1
deci x∈ (-1;8/7] cu conditia ca 2x-1 sa se divida cu 3 deci 2x=3p+1, p ∈Z
 x= (3p+1)/2 functie crescatoare de p si practic x∈Q
pt p=-1, x=-1∉(-1;8/7]
pt p=0, x=1/2∈(-1;8/7]
pt p=1, x=2∉(-1;8/7]
deci singura solutie x=1/2

verificare[3/10]=0 adevarat



Extra,
 cum le faceam eu pana acum

[(x+1)/5]=(2x-1)/3
(2x-1)/3 ∈Z,
2x-1=3p, p∈Z
2x=3p+1
x=(3p+1)/2
discutie functie de x

p=0, x=1/2 ,  (2x-1)/3=0
[(1+1/2)/5]=[(3/2)/5]=0 verifica x=1/2 deci solutie
p=1, 2x=4, x=2,(2x-1)/3=1
p=2,2x=7, x=7/2  (2x-1)/3=6/3=2

[(1+7/2)/5]=2
[(9/2)/5]=2 fals, nu verifica
apoi , pt valori cescatoare ale lui x, panta 2/3 fiind mai mare decat panta 1/5 cu care creste intrepte functia [(x+1)/5], drepta (2x-1)/3 se va indeparta de zona  treptelor functiei [(x+1)/5]
deci vom cauta solutii pt p<0 si x<0
p=-1 , x= (-3+1)/2=-1, (2x-1)/3=-3/3=-1
[0/5]=-1
0=-1 fals nu verifica  ne asteptam ca si incontinuarea, ptx x<-1 sa nu mai avem solutii, datorita pantelor diferite ac al;e celor doua drepte
 p=-2, x=(-6+1)/2=-5/2...(2x-1)/3=-6/3=-2
[(-5/2+1)/5]=[(-3/2)/5]=[-3/10]=-1=-2 , fals

deci singura solutie x=1/2

Answer 2

[tex]\text{De fapt paranteza patrata inseamna partea intreaga.}\\ \left[\dfrac{x+1}{5}\right]=\dfrac{2x-1}{3}\\ Notam: \dfrac{2x-1}{3}=k\Rightarrow x=\dfrac{3k+1}{2},k\in \mathbb{Z}\\ \text{Stim ca:}\\ k\leq \dfrac{x+1}{5} \ \textless \ k+1|\cdot 5\\ 5k\leq \dfrac{3k+1}{2}+1\ \textless \ 5k+5|\cdot 2\\ 10k\leq3k+3\ \textless \ 10k+10|-10k\\ 0\leq 3-7k\ \textless \ 10|-3\\ -3\leq -7k\ \textless \ 7\\ \dfrac{3}{7}\geq k\ \textgreater \ -1\\ \text{Si cum:} k\in\mathbb{Z}\Rightarrow k=0\\ S:\boxed{x=\dfrac{1}{2}}[/tex]