Question

Stabiliti daca fractia F se simplifica cu 17 iar n ar fi numar natural.​

Answer 1

Răspuns:

Explicație pas cu pas:

(3ⁿ⁺¹•5ⁿ+3ⁿ•5ⁿ⁺²+6•3ⁿ•5ⁿ) : (2²ⁿ⁺¹•3ⁿ+3ⁿ⁺¹•4ⁿ+2ⁿ⁺¹•6ⁿ⁺¹) =

= (3•3ⁿ•5ⁿ+3ⁿ•5²•5ⁿ+6•3ⁿ•5ⁿ):(2•2²ⁿ•3ⁿ+3•3ⁿ•2²ⁿ+2•2ⁿ•6•6ⁿ) =

= [3ⁿ•5ⁿ•(3+25+6)] : (2•2²ⁿ•3ⁿ+3•3ⁿ•2²ⁿ+12•2ⁿ•2ⁿ•3ⁿ) =

=  (3ⁿ•5ⁿ•34) : (2•2²ⁿ•3ⁿ+3•3ⁿ•2²ⁿ+12•2²ⁿ•3ⁿ) =

=  (3ⁿ•5ⁿ•34) : [2²ⁿ•3ⁿ•(2+3+12)] =

=  (3ⁿ•5ⁿ•34) : (2²ⁿ•3ⁿ•17) =

= (2•5ⁿ):(2²ⁿ) ; fractia se simplifica prin 17•3ⁿ  ; n ∈ N