Question

Please help me with this math

Answer 1

Answer:

(E) 9

Step-by-step explanation:

$\frac{(3^{2008})^2-(3^{2006})^2}{(3^{2007})^2-(3^{2005})^2} = \\\frac{(3^{2007+1})^2-(3^{2007-1})^2}{(3^{2007})^2-(3^{2007-2})^2} = \\\frac{(3^{2007})^2(3^{2}-3^{-2} )}{(3^{2007})^2(1-3^{-4} )} =\\\frac{9-\frac{1}{9} }{1-\frac{1}{81} }=\\\frac{80}{9}:\frac{80}{81} = \\\frac{80}{9}*\frac{81}{80} = \frac{81}{9} = 9$