A=1+5+5^2+..+5^9/1+5+5^2+...+5^8

=1+5^9/1+5+5^2+...+5^8 

B=1+3+3^2+..+3^9/1+3+3^2+..+3^8

=1+3^9/1+3+3^2+..+3^8

đặt A' =1+5+5^2+...+5^8

5A'=5+5^2+5^3+...+5^9

5A'-A'=5+5^2+5^3+...+5^9-5-1-5-5^2-...-5^8

4A'=5^9-1=>A'=(5^9-1):4

tương tự B'=(3^9-1):4

A=1+5^9/(5^9-1)/4=4.5^9/5^9-1

B=1+3^9/(3^9-1)/4=4.3^9/3^9-1

=> A<B

mk nghĩ là A>B

\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)

\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)

\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)  

Có: \(\frac{1}{1+5+5^2+...+5^8}>0\)              và      \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)

\(\Rightarrow A>B\)

Ta đặt  \(A=1+5+5^2+......+5^9\Rightarrow5A=5+5^2+...+5^9+5^{10}\)

\(\Rightarrow4A=5^{10}-1\Rightarrow A=\frac{5^{10}-1}{4}\)

tTương tự \(B=1+5+5^2+......+5^8\Rightarrow B=\frac{5^9-1}{4}\)

\(C=1+3+3^2+......+3^9\Rightarrow C=\frac{3^{10}-1}{3}\)

\(D=1+3+3^2+......+3^8\Rightarrow D=\frac{3^9-1}{3}\)

Vậy \(\frac{A}{B}=\frac{5^{10}-1}{5^9-1}=\frac{5\left(5^9-1\right)+4}{5^9-1}=5+\frac{4}{5^9-1}\)

\(\frac{C}{D}=\frac{3^{10}-1}{3^9-1}=\frac{3\left(3^9-1\right)+3}{3^9-1}=3+\frac{3}{3^9-1}\)

Ta thấy \(\frac{3}{3^9-1}< 1\Rightarrow3+\frac{3}{3^9-1}< 4< 5< 5+\frac{5}{5^9-1}\)

Vậy \(\frac{A}{B}>\frac{C}{D}\)  hay \(\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}>\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)

ta có: \(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^9}=1\)

mà \(1+3+3^2+...+3^9>1+3+3^2+...+3^8\)

\(\Rightarrow B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}>1\)

\(\Rightarrow A< B\)

Ta thấy : A= ( 1+5+5^2+.......+5^9)/(1+5+5^2+...... +5^8)= 5^9

B=(1+3+3^2+......+3^9)/(1+3+3^2+,,,,,,,,+3/9)=1

mÀ 5^9 > 1 . SUY RA  A>B

Vậy A>B

mk ko chắc chắn lắm 

k cho mk nhé