Question

Ma ajutati va rog cei care stiu!

Answer 1

^ la puterea
[2^(n+3)+2^(n+2)+2^(n+1)+2^n] / [3^(n+2)+2*3^(n+1)+3^n] =
=[2^n)*(2³+2²+2+1] / [3^n(3²+2*3+1] =
=[2^n*(8+4+2+1] / [3^n(9+6+1] =
=2^n*16 / 3^n *16
se divide cu 16  (nu cu 6)

[2^(2n+1)+4^(n+1)-5*4^n] / [3^(n+1)-2*3^n] =
=[2^(2n+1)+2^(2n+2)-5*2^2n] / [3^(n+1)-2*3^n] =
 =[2^2n*(2+2²-5] / [3^n*(3-2)] =
 =2^2n / 3^n deci ireductibil