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 \(\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}>\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)

a) \(\left(\frac{2}{3}+\frac{1}{5}\right)^2:\left(\frac{2}{5}-\frac{1}{3}\right)\)

\(=\left(\frac{13}{15}^2\right)\cdot15\)

\(=\frac{169\cdot15}{225}\)

\(=\frac{169}{15}\)

b) 

\(\left(2\frac{1}{3}-1\frac{3}{5}\right)\cdot\left(2\frac{4}{9}:3\frac{1}{2}\right)^2\)

\(=\left(\frac{7}{3}-\frac{8}{5}\right)\cdot\left(\frac{22}{9}\cdot\frac{7}{2}\right)^2\)

\(=\frac{11\cdot5929}{15\cdot81}\)

\(=53,6781893\)