Bagaimana cara termudah untuk mencari.... Question from @azzammuddin911

October 2018 2 36 Report

Bagaimana cara termudah untuk mencari $\frac{3^2^0^0^8(10^2^0^1^3+5^2^0^1^22^2^0^1^1)}{5^2^0^1^2(6^2^0^1^0+3^2^0^0^92^2^0^0^8}$

ImmanuelSinaga $$\begin{align} \frac{3^{2008}(10^{2013}+5^{2012}2^{2011})}{5^{2012}(6^{2010}+3^{2009}2^{2008})}~~&=~~\frac{3^{2008}(10^{2013}+10^{2011}~\times5)}{5^{2012}(6^{2010}+6^{2008}\times3)} \\ &=~~\frac{3^{2008}10^{2011}(10^2+5)}{5^{2012}6^{2008}(6^2+3)} \\ &=~~\frac{3^{2008}2^{2011}5^{2011}(100+5)}{5^{2012}2^{2008}3^{2008}(36+3)} \\ &=~~\text{Bagikan saja} \\ &=~~\frac{2^3\times105}{5\times39} \\ &=~~\frac{56}{13} \end{align}$

3 votes Thanks 2

ydhh $\frac{3^{2008}(10^{2013}+5^{2012}2^{2011})}{5^{2012}(6^{2010}+3^{2009}2^{2008})}=\frac{3^{2008}(100\cdot 10^{2011}+5\cdot(5^{2011}2^{2011}))}{5^{2012}(36\cdot 6^{2008}+3\cdot (3^{2008}2^{2008}))} \\ =\frac{3^{2008}((100+5)10^{2011})}{5^{2012}((36+3)6^{2008})} \\ =\frac{105\cdot 3^{2008}\cdot 5^{2011}\cdot 2^{2011}}{39\cdot 5^{2012}\cdot 2^{2008}\cdot 3^{2008}} \\ =\frac{105\cdot 2^3}{39\cdot 5} \\ =\frac{56}{13}$

1 votes Thanks 1

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