Cho hai biểu thức A = (1 - 1/2) (1 - 1/3) (1 - 1/4) ... (1 - 1/19) (1 - 1/20)

* Hướng dẫn giải

a) Ta có $A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\mathrm{...}\frac{18}{19}.\frac{19}{20}=\frac{1}{20}>\frac{1}{21}$

b) Ta có: 

$$\begin{gathered} B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}\mathrm{...}\frac{80}{81}.\frac{99}{100} \\ =\frac{3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}\mathrm{...}\frac{8.10}{9^2}.\frac{9.11}{{10}^2} \\ =\frac{{2.3}^2{.4}^2{.5}^2{\mathrm{...9}}^2\mathrm{.10.11}}{2^2{.3}^2{.4}^2{.5}^2{\mathrm{...9}}^2{.10}^2} \\ =\frac{11}{20}>\frac{11}{21} \end{gathered}$$

Vậy $B>\frac{11}{21}$