Răspuns:
x∈[2/3; 4/5)
Explicație pas cu pas:
[x]∈Z
posibil
doar 0+1 sau 1+0 ***
cum x+1/5<x+1/3
convine doar =[x+1/3]=1...x∈[2/3, 5/3)
SI
[x+1/5]=0....x∈[0,4/5)
[2/3, 5/3)∩[0,4/5)=[10/15, 25/15)∩[0, 12/15)=[10/15; 12/15)=[2/3; 4/5)
GREA! (pt mine...:) )
***P.S .
nu am verificat si
1=2+(-1)
[x+1/3]=2...x∈[2/3; 8/3)
[x+1/5]=-1...x∈[-6/5;-1/5) cu intersectie vida
analog si urmatoarele perechi posibile
deci ramane un singur interval
