Răspuns:
16^13=2^52
8^17=2^51
2^51>2^50. rezulta ca modulul se desface astfel:
2^52+2^51-2^50=2^50(2^2+2^1-1)=2^50(4+2-1)=2^50×5
[tex]\it 16^{13} =(2^4)^{13}=2^{52}\\ \\ 8^{17}=(2^3)^{17}=2^{51}\\ \\ |2^{50}-8^{17}|=|2^{50}-2^{51}|=|2^{51}-2^{50}|=|2\cdot2^{50}-2^{50}|=|2^{50}|=2^{50}[/tex]
Exercițiul devine:
[tex]\it 2^{52}+2^{50}=2^2\cdot2^{50}+2^{50} =5\cdot2^{50}[/tex]
