Răspuns:
Explicație pas cu pas:
a) 3x=2, x=2/3
b) x∈Ф
c) x∈(-3,3)
d) |3x-4|≤5
-5≤3x-4≤5 |+4
-1≤3x≤9 :3
-1/3≤x≤3
x∈[-1/3,3]
e) y ∈ R- {2}
f) a∈R
g) x∈R
4. a) (3n+1-3n)/n >1/5
1/n > 1/5
n<5
n∈(-∞,0)∪(0,5)
b) [2(n+1)-(2n+3)]/2(2n+3) >1/12
2n+2-2n-3/2(2n+3)>1/12
-1/2(2n+3)>1/12
*(-2)
1/2n+3<-1/6
2n+3>-6
2n>-9
n>-9/2
n∈Z, n={0,1,2,3....} adica n ∈N
c) 1-2/(n^2+1)≥1/10
-1/(n^2+1)≥1/10
1/(n^2+1)≤-1/10
n^2+1≥-10
n∈Z
