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

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

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

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

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

=\(\frac{1}{80}\)<\(\frac{1}{9}\)

Vậy tổng trên nhỏ hơn \(\frac{1}{9}\)

Ta có \(\dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{80}\right)=\dfrac{1}{3}.\dfrac{3}{80}=\dfrac{1}{80}< \dfrac{1}{9}\)

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 \(\dfrac{1}{80}< \dfrac{1}{9}\)

Đặt vế trái là B

\(3B=\frac{23-20}{20.23}+\frac{26-23}{23.26}+\frac{29-26}{26.29}+...+\frac{80-77}{77.80}\)

\(3B=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}=\frac{1}{20}-\frac{1}{80}\)

\(3B=\frac{3}{80}\Rightarrow B=\frac{1}{80}< \frac{1}{9}\)

Ta có: \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)

\(=\frac{1}{3}\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)

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

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

\(=\frac{1}{3}.\frac{3}{80}=\frac{1}{80}< \frac{1}{9}\)

Vậy \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\)

:3 B<1/79 mà 1/79<1/9=>B<1/9

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

= \(\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)

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

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

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

= \(\frac{1}{80}\) < \(\frac{1}{9}\)

⇒ \(\frac{1}{20.23} +\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\) < \(\frac{1}{9}\) (ĐPCM)

ko bt sory nha

Ta có:

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

\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)

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

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

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

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

Mà \(\frac{1}{80}< \frac{1}{9}\)

\(\Rightarrow\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\left(đpcm\right)\)

Học tốt 

CMR:\(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+....+\frac{1}{77.80}< \)\(\frac{1}{9}\)

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

Khi đó ta có:

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

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

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

So sánh:  \(\frac{1}{80}< \frac{1}{9}\)(ĐPCM)

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

=>A=\(\frac{1}{3}\).(\(\frac{3}{20.23}+\frac{3}{23.26}+....+\frac{3}{77.80}\))

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

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

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

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

Do \(\frac{1}{80}\)<\(\frac{1}{9}\)

Nên \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+....+\frac{1}{77.80}< \frac{1}{9}\)

ko bt

giúp mik nha đng cần gấp

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

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

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

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

\(=\frac{1}{80}< \frac{1}{9}\)

Ta có: \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)

= \(\frac{1}{3.}\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)

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

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

= \(\frac{1}{3}.\frac{3}{80}=\frac{1}{80}< \frac{1}{9}\)

Ta có B= 3(\(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\) )

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

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

        B=3.\(\frac{3}{80}\)=\(\frac{9}{80}\)

\(\frac{3}{9}B=\frac{3}{9}.\left(\frac{9}{20.23}+\frac{9}{23.26}+\frac{9}{26.29}+...+\frac{9}{77.80}\right)\)

=> \(\frac{3}{9}B=\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+....+\frac{3}{77.80}\)

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

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

=> \(B=\frac{3}{80}:\frac{3}{9}=\frac{3}{80}.\frac{9}{3}=\frac{9}{80}\)