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

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

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

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

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

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

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

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

-.-

k rùi biết

\(\dfrac{3^2}{20.23}+\dfrac{3^2}{23.26}+\dfrac{3^2}{26.29}+...+\dfrac{3^2}{77.80}\)

\(=3\left(\dfrac{3}{20.23}+\dfrac{3}{23.26}+\dfrac{3}{26.29}+...+\dfrac{3}{77.80}\right)\)

\(=3\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{29}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)\)

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

\(=3\left(\dfrac{4}{80}-\dfrac{1}{80}\right)=3.\dfrac{3}{80}=\dfrac{9}{80}\)

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 )

Mọi người tk mình đi mình đang bị âm nè!!!!!!

Ai tk mình mình tk lại nha !!!

Ban giai di xong minh k cho nhe

#)Giải:

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

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

     \(A=\frac{1}{20}-\frac{1}{80}\)

     \(A=\frac{3}{80}< \frac{1}{9}\)

\(\Leftrightarrow A< \frac{1}{9}\)

          #~Will~be~Pens~#

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

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

\(=\frac{1}{3}.\frac{3}{80}\)

\(=\frac{3}{240}=\frac{1}{80}\)

Vì \(\frac{1}{80}< \frac{1}{9}\)

Nên \(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}< \frac{1}{9}\)

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

\(\frac{A}{3}=\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+\frac{3}{26\cdot29}+...+\frac{3}{77\cdot80}\)

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

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

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

\(A=\frac{3}{80}:3=\frac{3}{240}=\frac{1}{80}\)

\(\Rightarrow A=\frac{1}{80}< \frac{9}{720}< \frac{80}{720}=\frac{1}{9}\)

Vậy A < 1/9