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\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
\(A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)
Đặt: \(B=2^{2018}+2^{2017}+2^{2016}+....+2^1+2^0\)
\(\Rightarrow2B=\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)
\(\Rightarrow2B-B=\left(2^{2019}+2^{2018}+2^{2017}+...+2^2+2\right)-\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)
\(\Rightarrow B=2^{2019}-1\)
\(\Rightarrow A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)
\(=2^{2019}-\left(2^{2019}-1\right)=2^{2019}+2^{2019}+1>1\)
\(3a=3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)
\(2a=3a-a=3-\frac{1}{3}-\frac{1}{3^{2019}}< 3\Rightarrow a< \frac{3}{2}\)