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29 tháng 4 2015

3^2= 9 
Vậy thì sẽ là:
9/ 20.23+ 9/ 23.26+...9/77.80
cách nhau 3 bỏ 3 ra ngoài
= 3(3/20.23+...3/77.80)
=3(3/20-3/23+3/23-3/26+.....+3/77-3/80)
=3(3/20-3/80)
=3. 9/80
=27/80<1

27 tháng 4 2017

32=9

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

=\(\frac{9}{20.23}\)+\(\frac{9}{23.26}\)+...+\(\frac{9}{77.80}\)

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

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

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

=3(\(\frac{4}{80}-\frac{1}{80}\))

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

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

Vậy\(\frac{9}{80}< 1\)

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

1 tháng 5 2017

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

=> \(\dfrac{9}{20.23}+...+\dfrac{9}{77.80}\)

= 9.\(\left(\dfrac{1}{20.23}+...+\dfrac{1}{77.80}\right)\)

\(=9.\left(\dfrac{1}{20.3}-\dfrac{1}{23.3}+\dfrac{1}{23.3}-\dfrac{1}{26.3}+...+\dfrac{1}{77.3}-\dfrac{1}{80.3}\right)\)= \(9.\left(\dfrac{1}{20.3}-\dfrac{1}{80.3}\right)\)

\(=9.\dfrac{1}{80}\)=\(\dfrac{9}{80}=0,1125< 1.\)


1 tháng 5 2017

cảm ơn bn nha

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

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

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

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

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

12 tháng 5 2015

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

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

\(=3\left(\frac{4}{80}-\frac{1}{80}\right)\)

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

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

12 tháng 5 2015

Katherine Lilly Filbert đúng rồi

25 tháng 4 2017

Đặt A= ...(như trên)

=>\(\dfrac{1}{3}A=\dfrac{1}{3}.\left(\dfrac{3^2}{20.23}+\dfrac{3^2}{23.26}+...+\dfrac{3^2}{77.80}\right)\)

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

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

=>\(\dfrac{1}{3}A=\dfrac{1}{20}-\dfrac{1}{80}\\ =>\dfrac{1}{3}A=\dfrac{4}{80}-\dfrac{1}{80}\\ =>\dfrac{1}{3}A=\dfrac{3}{80}=>A=\dfrac{3}{80}:\dfrac{1}{3}\\ =>A=\dfrac{3}{80}.3=\dfrac{9}{80}< 1\)

Vậy A<1 . Chúc bạn học tốt ! :)