{x}=(x-1)/2
C.E.
0≤(x-1)/2<1
0≤x-1<>2
1≤x<3⇒[x]∈{1;2}
{x}=x-[x]
(x-1)/2=x-1⇒(x-1)/2=0⇒x=1 care ∈[1;3) siverifica {1}=0
(x-1)/2=x-2⇒2x-4=x-1....x=3 care nu apartine dom de def...
deci solutie unica x=1
x/2∈[0,1)⇒x∈[0;2)⇒[x]∈{0;1}
{3x}=3x-[3x]
x/2=3x-[3x]
x/2=3x-0
0=5x/2...x=0care apartine [0.2)- si verifica {3*0}=0/2
x/2=3x-1
1=3x-x/2
1=5x/2
x=2/5 care ∈[0,2) si vweerifica {3*2/5}={6/5}=1/5 si (2/5) :2=1/5
