Cho 

- Hướng dẫn giải

$A=\left(40+\frac{3}{8}+\frac{5}{8^3}\right)+\left(\frac{7}{8^2}+\frac{4}{8^4}\right)$

$\begin{array}{l}B=\frac{3}{8}+40+\frac{5}{8^2}+\frac{5}{8^3}+\frac{5}{8^4}\\ \\ =\left(40+\frac{3}{8}+\frac{5}{8^3}\right)+\left(\frac{5}{8^2}+\frac{5}{8^4}\right)\end{array}$

$\begin{array}{l}Mà\frac{7}{8^2}+\frac{4}{8^4}=\left(\frac{5}{8^2}+\frac{4}{8^4}\right)+\frac{2}{8^2}; \\ \\ \frac{5}{8^2}+\frac{5}{8^4}=\frac{5}{8^2}+\frac{4}{8^4}+\frac{1}{8^4}\end{array}$

$Do \frac{2}{8^4}>\frac{1}{8^4}=>\frac{7}{8^2}+\frac{4}{8^4}>\frac{5}{8^2}+\frac{5}{8^4}=>A>B$