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

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

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

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

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

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

Vậy tổng trên nhỏ hơn \(\frac{1}{9}\)

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

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 \(\dfrac{1}{80}< \dfrac{1}{9}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

oh oh oh oh

Bạn ơi cách này dễ hiểu hơn nè:

Đặt X =\(\frac{1}{20.23}\)+\(\frac{1}{23.26}\)+..................+\(\frac{1}{77.80}\).Ta có:

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

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

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

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

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

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

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

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

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

Chúc bạn học tốt!

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

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