Question

Sa se determine Mulțimea :
A={ x aparține lui Z / exista n aparține lui Z astfel încât x= (6n+18) : 2n + 1 }

Answer 1

x= 3(2n+6) : 2n+1

2n+1/2n+6
2n+1/2n+1 - (se scad) ⇒2n+1/2n+6-2n-1⇒ 2n+1/5⇒
⇒2n+1∈ {1,5,-1,-5}⇒ 2n ∈ {0,4,-2,-6} ⇒ n ∈ {0,2,-1,-3}
n=0 ⇒ x= 3(2*0+6) : (2*0+1)=3*6:1=18
n=2 ⇒ x= 3(2*2+6) : (2*2+1) =3*10:5=30:5=6
n=-1⇒ x= 3[2*(-1)+6] : [2*(-1)+1]=12:(-1)=-12 
n=-3⇒ x= 3[2*(-3)+6] : [2*(-3)+1]= 0
x∈ {-12 ,0 ,6,18}
A={-12,0,6,18}