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\(\frac{2^{2010}\:+\:2^{2008}}{2^{2005}\:-\:2^{2003}}\)

 

For some reason, the answer 5/3 is not correct... anyone has some thoughts on their answer?

 Jun 17, 2019

Best Answer 

 #1
avatar+8778 
+2

\(\ \phantom{=\quad}\dfrac{2^{2010}\:+\:2^{2008}}{2^{2005}\:-\:2^{2003}}\\~\\~\\ =\quad\dfrac{2^{2008}(2^{2}\:+\:1)}{2^{2003}(2^{2}\:-\:1)}\\~\\~\\ =\quad\dfrac{2^{2008}}{2^{2003}}\cdot\dfrac{2^{2}+1}{2^{2}-1}\\~\\~\\ =\quad2^{5}\cdot\dfrac{2^{2}+1}{2^{2}-1}\\~\\~\\ =\quad2^{5}\cdot\dfrac{5}{3}\\~\\~\\ =\quad32\cdot\dfrac{5}{3}\\~\\~\\ =\quad\dfrac{160}{3}\)_

 Jun 17, 2019
 #1
avatar+8778 
+2
Best Answer

\(\ \phantom{=\quad}\dfrac{2^{2010}\:+\:2^{2008}}{2^{2005}\:-\:2^{2003}}\\~\\~\\ =\quad\dfrac{2^{2008}(2^{2}\:+\:1)}{2^{2003}(2^{2}\:-\:1)}\\~\\~\\ =\quad\dfrac{2^{2008}}{2^{2003}}\cdot\dfrac{2^{2}+1}{2^{2}-1}\\~\\~\\ =\quad2^{5}\cdot\dfrac{2^{2}+1}{2^{2}-1}\\~\\~\\ =\quad2^{5}\cdot\dfrac{5}{3}\\~\\~\\ =\quad32\cdot\dfrac{5}{3}\\~\\~\\ =\quad\dfrac{160}{3}\)_

hectictar Jun 17, 2019

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