Question

simplificati fractiile 2^n *3^(n+1)+6^(n+1)/2^(n+1)*3^(n+1)+6^(n+1)

Answer 1

[2^n *3^(n+1)+6^(n+1)]/[2^(n+1)*3^(n+1)+6^(n+1)]=

Mai intai numaratorul:

2^n *3^(n+1)+6^(n+1)=2^n *3^n *3 +2*2^n *3*3^n=
=(2^n *3^n)(3+6)=9*(2^n)*(3^n)

Apoi numitorul:

2^(n+1)*3^(n+1)+6^(n+1)=2^(n+1)*3^(n+1)+2^(n+1)*3^(n+1)=
=2*2^(n+1)*3^(n+1)=2*2*3*(2^n)*(3^n)=12*(2^n)*(3^n)

Fractia devine:

[9*(2^n)*(3^n)]/[12*(2^n)*(3^n)]=9/12=3/4