Question

Mă poate ajuta cineva cu calculul "(2^3n+2 ÷ 8^n)^2 + (3^2n+1 × 8^n) : (9^n × 2^3n) unde n este număr natural??

Answer 1

[(2^(3n+2) : 8^n]^2 + [3^(2n + 1) × 8^n] : [9^n × 2^(3n)] =

= [(2^(3n+2) : (2^3)^n]^2 + [3^(2n+1) × (2^3)^n] : [(3^2)^n × 2^(3n)] =

= [(2^(3n+2) : 2^(3n)]^2 + [3^(2n+1) × 2^(3n)] : [3^(2n) × 2^(3n)] =

= (2^2)^2 + (3 × 1) =

= 2^4 + 3 =

= 16 + 3 =

= 19