Question

SA SE ARATE CA FRACTIA 3 la puterea n+1 · 2 la puterea2 n +4 la puterea n+1 ·3la n +3la n ·2la 2n +1 SUPRA 2la 2n · 3la n +12 la n+1 -3la n +1 ·4la n +1 ES
TE NUMAR NATURAL , ORICARE AR FI n ∈ N

Answer 1

[3^(n+1)·2^2n+4^(n+1)·3^n+3^n·2^(2n+1)]/[2^2n·3^n+12^(n+1)-3^(n+1)·4^(n+1)]=
[3^n·3·2²·2^n+4^n·4·3^n+3^n·2²·2^n·2]/[2²·2^n·3^n+12^n·12-3^n·3·4^n·4]=
[6^n·12+2^2n·4·3^n+6^n·8]/[4·6^n+(3·2²)^n·12-3^n·2^2n·12]
[6^n·12+2²·2^n·4·3^n+6^n·8]/[4·6^n+3^n·2²·2^n-3^n·2^n·2²·12]=
[6^n·12+16·6^n+6^n·8]/[4·6^n+4·6^n-48·6^n]=
[6^n(12+16+8)]/[5^n(4+4+48)]= simplificam cu 6^n
36/56=simplificam cu 4=9/14