Question

Determinati k apartine z pentru care numerele x= k+3\5 si y = 3k-1\5 sunt simultan intregi.
VA ROG ESTE URGENT SI VA MAI ROG DATI-MI REZOLVAREA COMPLETA...Multumesc !!!

Answer 1

(k + 3 ) ∈ D5   k+3 ∈ {-5; -1; 1; 5}    k ∈ {-8; -4; -2; 2}
(3k - 1) ∈ D5   3k-1 ∈ (-5; -1; 1; 5;}  3k ∈ {-4; 0; 2; 6}     k ∈ {0; 2}
{-8; -4; -2; 2} ∩ {0; 2}  = 2   k = 2

Answer 2

x   si   y    trebuie  sa   fie   multiplii   de   5
Pt   k>0
U(x)={0,5}
U(k+3)=0=>k1={7,17,27...}
U(k+3)=5  =>k2={2,12,22,...}
Pui   aceleasi   conditii   pt  U(y)
U(3k-1)=0  K3={7,17,27...}
U(3k-1)=5 =>  U(3k)=6  => k4={2,12,22...}
Determini  
k1Uk2   si   K3 UK4   Le   iyersectezi   si   obtii
[K1Uk2]∩[K3UK4]={7,17,27...}U{2,12,22,...} (A
revin  imediat
Cazul   k<0
U(k+3)=5  => k5={-8,-18,...]
U(k+3)=0 => k6={-3,-13,-23,...}
U(3k-1)=0  k7={-3,-13,-23....}
U(3k-1)=5  =<  k8=(-8,-18,-28...}
[k5Uk6]∩{k7UK8[={-8,-18,-28,...}U{-3.-13,-23...} B{
In   final   faci  AUB