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

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

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

\(=\frac{1}{3}.5\)

\(=\frac{5}{3}\)

\(\Rightarrow A>\frac{5}{3}>\frac{5}{4}\)

\(\Rightarrow A>\frac{5}{4}\)

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

\(\Rightarrow1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}>\frac{9}{4}\)

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

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

Đề ??? :

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

Giải

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

\(\Rightarrow A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+...+\frac{1}{27}\right)+\left(\frac{1}{29}+...+\frac{1}{81}\right)+\left(\frac{1}{83}+...+\frac{1}{243}\right)\)

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

\(=\frac{1}{3}.5\)

\(=\frac{5}{3}\)

\(\Rightarrow A>\frac{5}{3}>\frac{5}{4}\)

\(\Rightarrow A>\frac{5}{4}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{5}+...+\frac{1}{397}>\frac{9}{4}\)

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

\(\Rightarrow\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}>\frac{9}{20}\left(đpcm\right)\)

forever alone rai đề rùi

sao anh đăng lắm câu hỏi vậy ?

Bài anh có gì em tham khảo nhé 

Là sao??? Lê Tài Bảo Châu

là ních phụ của Lê Tài Bảo Châu 

Đợi anh cái lúc đó nhìn nhầm đề bài :))