Question

3power2005+3power2003/3power2004-3power2002

Answer 1

Step-by-step explanation:

$\frac{ {3}^{2005} + {3}^{2003} }{ {3}^{2004} - {3}^{2002} }$

$= \frac{ {3}^{2002}( {3}^{2} + {3}^{1} ) }{ {3}^{2002} ( {3}^{2} - 1)}$

$= \frac{27 + 3}{9 - 1}$

$= \frac{30}{8}$

$= \frac{15}{4}$