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

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

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

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

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

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

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

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

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

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

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

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

Đặt A=3²/20.23+3²/23.26+^{....................}+3²/77.80

      A=3(3/20.23+3/23.26+.........+3/77.80)

     A=3(1/20-1/23+1/23-1/26+.+1/77-1/80)

     A=3(1/20-1/80)

    A=3.3/80

    A=9/80                       Mà A=9/80<1         =>A<1                   (đpcm)

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 ! :)

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

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

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

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

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

Katherine Lilly Filbert đúng rồi

1/3=3/20*23+3/23*26+...+3/77+80

1/3=1/20-1/23+1/23-1/26+...+1/77-1/80

1/3=1/20-1/80

1/3=3/80

-> 3/3=3/80*3

->9/80

Vì 9/80<1 nên: => 3^2/20*23+3^2/23*26+...+3^2/77*80

Ta có
\(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\)
\(A=3^2\left(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\right)\)
\(A=3^2\cdot\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(A=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(A=3\cdot\frac{3}{80}=\frac{9}{80}< 1\left(9< 80\right)\)

\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}<\frac{1}{8}\)

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

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

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

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

\(\Rightarrow\frac{9}{80}=\frac{1}{8}\)