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bai 3
\(A=\frac{10^{2004}+1}{10^{2005}+1}\)
\(10A=\frac{10^{2004}+10}{10^{2005}+1}\)
\(10A=1\frac{9}{10^{2005}+1}\)
\(B=\frac{10^{2005}+1}{10^{2006}+1}\)
\(10B=\frac{10^{2005}+10}{10^{2006}+1}\)
\(10B=1\frac{9}{10^{2006}+1}\)
Vì \(1\frac{9}{10^{2005}+1}>1\frac{9}{10^{2006}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
bai 4
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^8}\)
\(\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^9}\)
\(A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{3^9}\)
\(\frac{2004\times2004+3006}{2005\times2005-1003}=\frac{2004\times2004+3006}{\left(2004+1\right)\times\left(2004+1\right)-1003}\)
\(\frac{2004\times2004+3006}{2004\times\left(2004+1\right)+1\times\left(2004+1\right)-1003}=\frac{2004\times2004+3006}{2004\times2004+2\times2004+1-1003}\)
\(=\frac{2004\times2004+3006}{2004\times2004+3006}=1\)