Question

[Lê Minh Khánh]

cho biểu thúc A=\(\dfrac{3}{4}\)+\(\dfrac{8}{9}\)+\(\dfrac{15}{16}\)+....+\(\dfrac{9999}{10000}\)  chứng minh  A<99

Answer 1

[Vũ Nguyễn Nam Khánh]

Tham khảo :
 

$3.\genfrac{}{}{0ex}{}{8}{9}.\genfrac{}{}{0ex}{}{15}{16}.....\genfrac{}{}{0ex}{}{9999}{10000}$

$=\genfrac{}{}{0ex}{}{1.3}{2.2}.\genfrac{}{}{0ex}{}{2.4}{3.3}.\genfrac{}{}{0ex}{}{3.5}{4.4}.....\genfrac{}{}{0ex}{}{99.101}{100.100}$

$=\genfrac{}{}{0ex}{}{\left(\mathrm{1.2.3.....99}\right)}{\left(\mathrm{2.3.4.....100}\right)}.\genfrac{}{}{0ex}{}{\left(\mathrm{3.4.5.....101}\right)}{\left(\mathrm{2.3.4.....100}\right)}$

$=\genfrac{}{}{0ex}{}{1}{100}.\genfrac{}{}{0ex}{}{101}{2}=\genfrac{}{}{0ex}{}{101}{200}$
 

Answer 2

[Vũ Nguyễn Nam Khánh]