Cho A = 40 + 3/8 + 7/8^2 + 5/8^3 + 32/8^5 và B = 24/8^2 + 40

* Hướng dẫn giải

$\begin{array}{l}\text{A-B=}40+\frac{3}{8}+\frac{7}{8^2}+\frac{5}{8^3}+\frac{32}{8^5}-\left(\frac{24}{8^2}+40+\frac{5}{8^2}+\frac{40}{8^4}+\frac{5}{8^4}\right)\\ =40+\frac{3}{8}+\frac{7}{8^2}+\frac{5}{8^3}+\frac{32}{8^5}-\frac{24}{8^2}-40-\frac{5}{8^2}-\frac{40}{8^4}-\frac{5}{8^4}\\ =\left(40-40\right)+\left(\frac{3}{8}-\frac{24}{8^2}\right)+\left(\frac{7}{8^2}-\frac{5}{8^2}\right)+\left(\frac{5}{8^3}-\frac{40}{8^4}\right)+\left(\frac{32}{8^5}-\frac{5}{8^4}\right)\\ =0+0+\frac{2}{8^2}+0+\left(\frac{-1}{8^4}\right)>0\end{array}$

Vậy A - B > 0

Suy ra A > B.