1. [x+2]+[x-3]-[x+4]=3
se foloseste proprietatea [x+n]=[x]+n
[x]+2+[x]-3+[x]-4=3
[x]-5=3, [x]=8, x∈[8,9)
2. [x]+[x+1/2]=[2x]
se verifica prin calcul direct, sau in identitatea lui Hermite se pune n=2
[x]+[x+1/n]+[[x+2/n]+...[x+n-1/n]=[nx]
n=2⇒[x]+[x+1/2]=[2x]
3.{x}=(x-1)/2
x-[x]=(x-1)/2⇒[x]=x-(x-1)/2=(x+1)/2
x-1<[x]≤x⇒x-1<(x+1)/2≤x
2x-2<x+1, x<3
x+1≤2x, -x≤-1, x≥1
x∈[1,3)
dar (x+1)/2∈Z, deci x=1
5. notam [x+3/4]=k, k∈Z
k≤(x+3)/4<k+1
4k≤x+3<4k+4I-3
4k-3≤x<4k+1 (*)
cum [(x+3)/4]=(x+4)/5 (din enunt), deducem ca (x+4)/5=k⇒x=5k-4 (**)
din relatia (*) si (**), obtinem
4k-3≤5k-4<4k+1 I+4
4k+1≤5k<4k+5
4k+1≤5k, -k≤-1, k≥1
5k<4k+5, k<5
k∈{1,2,3,4}
revenim la notatia (**) x=5k-4, inlocuim pe k si obtinem x∈{1,6,11,16}
4. cred ca deoarece x²+x+1 nu are solutii in R deoarece Δ<0, ecuatia are o singura solutie x=0
