Ta có \(\dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{80}\right)=\dfrac{1}{3}.\dfrac{3}{80}=\dfrac{1}{80}< \dfrac{1}{9}\)
Ta có:
\(\dfrac{1}{20.23}+\dfrac{1}{23.26}+\dfrac{1}{26.29}+...+\dfrac{1}{77.80}\)
=\(\dfrac{1}{3}\left[\left(\dfrac{1}{20}-\dfrac{1}{23}\right)+\left(\dfrac{1}{23}-\dfrac{1}{26}\right)+\left(\dfrac{1}{26}-\dfrac{1}{29}\right)+...+\left(\dfrac{1}{77}-\dfrac{1}{80}\right)\right]\)
= \(\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{80}\right)\)
=\(\dfrac{1}{3}.\dfrac{3}{80}\)
=\(\dfrac{1}{80}\)
Vì \(\dfrac{1}{80}\)>\(\dfrac{1}{9}\)
Nên \(\dfrac{1}{20.23}+\dfrac{1}{23.26}+\dfrac{1}{26.29}+...+\dfrac{1}{77.80}>\dfrac{1}{9}\)
Ta có:
\(\\ \dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}\\ =\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{80}\right)\\ =\dfrac{1}{3}\left(\dfrac{3}{80}-\dfrac{1}{80}\right)\\ =\dfrac{1}{3}.\dfrac{3}{80}\\ =\dfrac{1}{80}\\ \)
Vậy \(\dfrac{1}{80}< \dfrac{1}{9}\)
\(\dfrac{1}{20.23}+\dfrac{1}{23.26}+\dfrac{1}{26.29}+...+\dfrac{1}{77.80}\)
\(=\dfrac{1}{3}.\left(\dfrac{3}{20.23}+\dfrac{3}{23.26}+\dfrac{3}{26.29}+...+\dfrac{3}{77.80}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{29}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{1}{20}-\dfrac{1}{80}\right)\)
\(=\dfrac{1}{3}.\dfrac{3}{80}\)
\(=\dfrac{1}{80}< \dfrac{1}{9}\left(đpcm\right)\)