Question

Ajutooor!!! Maare nevoiieee!!!

Answer 1

   
[tex]\displaystyle \\ b) \\\\ \frac{2^{n+3}+2^{n+2}+2^{n+1}+2^{n}}{3^{n+2}+2\times 3^{n+1}+3^{n}} =\\ \text{dam factor comun la numarator: } 2^{n-1}~~\text{si la numitor: } 3^{n-1}. \\ \\ = \frac{2^{n-1}(2^4+2^3+2^2+2^1)}{3^{n-1}(3^3+2\times 3^2+3^1)}=\\\\= \frac{2^{n-1}(16+8+4+2)}{3^{n-1}(27+2\times 9+3)}=\\\\ = \frac{2^{n-1} \cdot 30}{3^{n-1} \cdot 48}= \frac{2^{n-1} \cdot 5 \cdot 6}{3^{n-1} \cdot 8 \cdot 6} =\ \textgreater \ ~se~simplifica~cu~6[/tex]


[tex]\displaystyle \\c)\\\\   \frac{4^n \cdot 5^{n+1} + 4^{n+1} \cdot 5^n}{3^n \cdot 7^{n+1} + 3^{n+1} \cdot 7^n -21^n}= \\ \\ =\frac{4^n \cdot 5^{n+1} + 4^{n+1} \cdot 5^n}{3^n \cdot 7^{n+1} + 3^{n+1} \cdot 7^n -3^n \cdot 7^n}= \\ \\ =\frac{4^n \cdot 5^n (5 + 4)}{3^n \cdot 7^n(7+ 3 -1)}= \frac{20^n \cdot 9}{21^n \cdot 9} \text{Se simplifica cu 9.}[/tex]


[tex]\displaystyle \\ d) \\ \\ \frac{6^{n+1} + 6^n}{5^{n+2}-6 \cdot 5^n}= \frac{6^n(6 + 1)}{5^n(25-6)}= \frac{6^n \cdot 7}{5^n \cdot 19} ~~\text{ (fractie ireductibila)}[/tex]