Ta có :
+) \(A=\dfrac{1+9+9^2+...+9^{2009}}{1+9+9^2+...+9^{2009}}+\dfrac{9^{2010}}{1+9+9^2+...+9^{2009}}\)
\(A=1+1:\dfrac{1+9+9^2+...+9^{2009}}{9^{2010}}\)
\(A=1+1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)\)
+) \(B=\dfrac{1+5+5^2+...+5^{2009}}{1+5+5^2+...+5^{2009}}+\dfrac{5^{2010}}{1+5+5^2+...+5^{2009}}\)
\(B=1+1:\dfrac{1+5+5^2+...+5^{2009}}{5^{2010}}\)
\(B=1+1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)
Vì \(\dfrac{1}{9^{2010}}< \dfrac{1}{5^{2010}}\)
\(\dfrac{1}{9^{2009}}< \dfrac{1}{5^{2009}}\) (ngoặc cả mấy cài so sánh này vào rôi mời suy ra nhé)
==-.
\(\dfrac{1}{9}< \dfrac{1}{5}\)
\(\)=> \(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}< \dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\)
=> \(1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)>1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)
=> \(1+1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)>1+1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)
Hay A > B
