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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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!

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

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

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

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

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

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

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

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

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

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

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

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

Vậy tổng trên nhỏ hơn \(\frac{1}{9}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+\frac{3^2}{26.29}+...+\frac{3^2}{77.80}\)

=\(\frac{3.3}{20.23}+\frac{3.3}{23.26}+\frac{3.3}{26.29}+...+\frac{3.3}{77.80}\)

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

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

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

Ta có:

\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+\frac{3^2}{26.29}+...+\frac{3^2}{77.80}=3\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)=3.\left(\frac{1}{20}-\frac{1}{80}\right)=3.\frac{3}{80}=\frac{9}{80}\)

Giống kết quả của mk thôi