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C.\(\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^6}{2^{5+}2^5}=\frac{4^6}{3^6}.\frac{6^6}{2^5+2^5}=\frac{24^6}{3^6.\left(2^5+2^5\right)}=\frac{8^6}{2^5.\left(1+1\right)}\)=\(\frac{8^6}{2^6}\)=4^6=4096
\(4^{11}.25^{11}\le2^n.5^n\le20^{12}.5^{12}\)
\(\Rightarrow\left(2^2\right)^{11}.\left(5^2\right)^{11}\le2^n.5^n\le\left(2^2.5\right)^{12}.5^{12}\)
\(\Rightarrow2^{22}.5^{22}\le2^n.5^n\le2^{24}.5^{24}\)
\(\Rightarrow\left(2.5\right)^{22}\le\left(2.5\right)^n\le\left(2.5\right)^{24}\)
\(\Rightarrow22\le n\le24\Rightarrow n\in\left\{22;23;24\right\}\left(n\in N\right)\)
Ta có 411.2511=(4.25)11=10011=1022
2n.5n=(2.5)n=10n
2012.512=(20.5)12=10012=1024
⇒ 1022≤10n<1024
⇒ 22≤n<24
⇒ nϵ {22;23}
Bài 2:
1: \(5^n+5^{n+2}=650\)
\(\Leftrightarrow5^n\cdot26=650\)
\(\Leftrightarrow5^n=25\)
hay x=2
2: \(32^{-n}\cdot16^n=1024\)
\(\Leftrightarrow\dfrac{1}{32^n}\cdot16^n=1024\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^n=1024\)
hay n=-10
13: \(9\cdot27^n=3^5\)
\(\Leftrightarrow3^{3n}=3^5:3^2=3^3\)
=>3n=3
hay n=1