Đặt : \(A=\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)

\(\Rightarrow3A=\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\)

\(\Rightarrow3A=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\)

\(\Rightarrow3A=\frac{1}{20}+\left(\frac{1}{23}-\frac{1}{23}\right)+\left(\frac{1}{26}-\frac{1}{26}\right)+...+\left(\frac{1}{77}-\frac{1}{77}\right)-\frac{1}{80}\)

\(\Rightarrow3A=\frac{1}{20}-\frac{1}{80}\)

\(\Rightarrow3A=\frac{3}{80}\)

\(\Rightarrow A=\frac{3}{80}:3\)

\(\Rightarrow A=\frac{1}{80}\)

Vì 80 > 79 nên \(\frac{1}{80}< \frac{1}{79}\)hay \(A< \frac{1}{79}\)

~ Hok tốt ~

\(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+\frac{1}{26\cdot29}+...+\frac{1}{77\cdot80}\)

\(< \frac{1}{3}\left[\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+\frac{3}{26\cdot29}+...+\frac{3}{77\cdot80}\right]\)

\(< \frac{1}{3}\left[\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right]\)

\(< \frac{1}{3}\left[\frac{1}{20}-\frac{1}{80}\right]\)

\(< \frac{1}{3}\left[\frac{4}{80}-\frac{1}{80}\right]\)

\(< \frac{1}{3}\cdot\frac{3}{80}=\frac{1}{80}< \frac{1}{79}(đpcm)\)

Ta có :

\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)

\(=\frac{1}{3}\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)

\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)

\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{80}\right)\)

\(=\frac{1}{3}.\frac{3}{80}\left(\frac{3}{80}< 1\right)\)

\(\Leftrightarrow\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}< \frac{1}{3}\left(đpcm\right)\)

\(M=\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77x80}\)

\(M=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\)

\(M=\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\)

\(\frac{3}{80}=\frac{3x9}{80x9}=\frac{27}{720};\frac{1}{9}=\frac{1x80}{9x80}=\frac{80}{720}\)

Vì \(\frac{27}{720}< \frac{80}{720}\Rightarrow\frac{3}{80}< \frac{1}{9}\Rightarrow M< \frac{1}{9}\)

          #~Will~be~Pens~#

Hoàng Nguyên Hiếu:Sai rồi nha bạn

\(\frac{1}{20.23}=\frac{1}{20}-\frac{1}{23}\Leftrightarrow23-20=1\)

-.-

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Nguyễn Quang Thành trả lời cụ thể đi

Ta có : \(\frac{3^2}{20\cdot23}+\frac{3^2}{23\cdot26}+...+\frac{3^2}{77\cdot80}=\frac{1}{3}\cdot\left(\frac{1}{20}-\frac{1}{80}\right)=\frac{1}{3}\cdot\frac{3}{80}=\frac{1}{80}< 1\)             ( đpcm )