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7 tháng 7 2020

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

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

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

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

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

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

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

30 tháng 5 2021

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

30 tháng 5 2021

người j đâu mà giỏi ghê lun á

14 tháng 6 2020

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

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

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

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

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

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

Vậy tổng trên nhỏ hơn \(\frac{1}{9}\)

1 tháng 8 2021
G
22 tháng 6 2020

Đặt vế trái là B

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

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

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

22 tháng 6 2020

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

22 tháng 6 2020

1/20.23+1/23.26+1/26.29+...+1/77.80 < 3/20.23+3/23.26+3/26.29+...+3/77.80=1/20-1/23+1/23-1/26+...+1/77-1/80

=1/20-1/80=3/80 < 1/9 = 3/27

6 tháng 5 2019

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

6 tháng 5 2019

ko bt

6 tháng 5 2019

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

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

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

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

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

6 tháng 5 2019

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

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

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

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

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

11 tháng 4 2017

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

11 tháng 4 2017

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

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

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

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

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

7 tháng 8 2019

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

\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{27.80}\right)\)

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

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

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

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

7 tháng 8 2019

oh oh oh oh

22 tháng 6 2020

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 \(\frac{1}{80}< \frac{1}{9}\)

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

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