b, \(A=\dfrac{1+5+5^2+...+5^8+5^9}{1+5+5^2+...+5^8}\)\(=\dfrac{1+5+5^2+...+5^8}{1+5+5^2+...+5^8}+\dfrac{5^9}{1+5+5^2+...+5^8}\)
\(=1+\dfrac{5^9}{1+5+5^2+...+5^8}\)
\(B=\dfrac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
\(=\dfrac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}+\dfrac{3^9}{1+3+3^2+...+3^8}\)
Đặt \(C=\dfrac{5^9}{1+5+5^2+...+5^9}\) ; \(D=\dfrac{3^9}{1+3+3^2+...+3^9}\)
Ta lại có: \(\dfrac{1}{C}=\dfrac{1+5+5^2+...+5^9}{5^9}\)
\(=\dfrac{1}{5^9}+\dfrac{5}{5^9}+\dfrac{5^2}{5^9}+...+\dfrac{5^9}{5^9}\)
\(=\dfrac{1}{5^9}+\dfrac{1}{5^8}+\dfrac{1}{5^7}+...+\dfrac{1}{5}\)
\(\dfrac{1}{D}=\dfrac{1+3+3^2+...+3^9}{3^9}\)
\(=\dfrac{1}{3^9}+\dfrac{3}{3^9}+\dfrac{3^2}{3^9}+...+\dfrac{3^9}{3^9}\)
\(=\dfrac{1}{3^9}+\dfrac{1}{3^8}+\dfrac{1}{3^7}+...+\dfrac{1}{3}\)
Vì \(\dfrac{1}{5^9}>\dfrac{1}{3^9};\dfrac{1}{5^8}>\dfrac{1}{3^8};-;\dfrac{1}{5}>\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{5^9}+\dfrac{1}{5^8}+-+\dfrac{1}{5}>\dfrac{1}{3^9}+\dfrac{1}{3^8}+...+\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{C}< \dfrac{1}{D}\Rightarrow C>D\Rightarrow1+C>1+D\)
Mà \(1+C=A;1+D=B\) \(\Rightarrow A>B\)
Vậy A>B
