Question

Să determinăm următoarele muţimi:

B∩N,unde B={105/2 ;106 /3;107 /4 ;…; n+104 /n+1 },n∈N
A= {x∈Z|6x/2x+1 ∈Z}

Answer 1

daca n+104/n+1= (n+1+103)/n+1= n+1/n+1  + 103/n+1 = 1+  103/n+1 ∈ N ⇒
103/n+1 ∈ N⇒ n+1∈ {1,103 } ⇒n=103-1=102
102+104/102+1=206/103=2⇒ B n N={2}

2x+1/6x            ⇒ 2x+1/6x
2x+1/2x+1 /*3   ⇒ 2x+1/6x+3 /-   ⇒ 2x+1/6x+3-6x⇒ 2x+1/3⇒ 
⇒2x+1∈ {-1,1,3,-3}⇒ 2x∈ {-2,0,2,-4} ⇒ x∈{-1,0,1,-2} ⇒ A={-1,0,1,-2}