Cho A = 4 + 3/7 + 4/7^2 + -441/7^6 + 27/7^5 + 1/7^6 và B = 147/7^3

* Hướng dẫn giải

$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)$

$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)$

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