Cho

- Hướng dẫn giải

$\begin{array}{l}A=4+\frac{3}{7}+\frac{4}{7^2}+\frac{-9}{7^4}+\frac{27}{7^5}+\frac{1}{7^6}\\ \\ =\left(4+\frac{3}{7}+\frac{4}{7^2}+\frac{27}{7^5}\right)+\left(\frac{-9}{7^4}+\frac{1}{7^6}\right)\end{array}$

$\begin{array}{l}B=\frac{3}{7}+4+\frac{5}{7^6}+\frac{4}{7^2}+\frac{27}{7^5}+\frac{-9}{7^9}\\ \\ =\left(4+\frac{3}{7}+\frac{4}{7^2}+\frac{27}{7^5}\right)+\left(\frac{5}{7^6}+\frac{-9}{7^9}\right)\end{array}$

$Do \frac{-9}{7^4}+\frac{1}{7^6}<\frac{5}{7^6}+\frac{-9}{7^9}=>A<B$