\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)

\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)

\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)

\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)

Chứng minh rằng:\(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+....+\frac{1}{1985}< \frac{9}{20}\)

mk làm thế này đúng ko mọi người

Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+......+\frac{1}{243}\)

\(A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+....+\frac{1}{27}\right)+\left(\frac{1}{29}+\frac{1}{31}+\frac{1}{33}+....+\frac{1}{81}\right)+\left(\frac{1}{83}+\frac{1}{85}+\frac{1}{87}+.....+\frac{1}{243}\right)\)

\(=>A>\frac{1}{3}+\frac{1}{9}.3+\frac{1}{27}.9+\frac{1}{81}.27+\frac{1}{243}.81=\frac{1}{3.5}=\frac{5}{3}\)

\(=>A>\frac{5}{3}>\frac{5}{4}=>A< \frac{5}{4}\)

\(=>\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+....+\frac{1}{397}< \frac{5}{4}\)

\(=>1+\frac{1}{3}+\frac{1}{7}+....+\frac{1}{397}< \frac{5}{4}\)

\(=>\frac{1}{5}.\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+....+\frac{1}{397}\right)< \frac{9}{4}.\frac{1}{5}\)

\(=>\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+......+\frac{1}{1985}< \frac{9}{20}\)

Tính nhah ---- giúp mik giải nâ các bn thank nhiều nhiều

a)\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}+\frac{1}{3}\)

b) \(\frac{\frac{1}{3}-\frac{1}{5}-\frac{1}{7}}{\frac{2}{3}-0,4-\frac{2}{7}}+\frac{\frac{3}{8}-\frac{3}{16}-\frac{3}{32}+\frac{3}{64}}{\frac{1}{4}-\frac{1}{8}-\frac{1}{16}+\frac{1}{32}}\)

c) \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)