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 ! :)

\(M=\dfrac{6}{10.13}+\dfrac{6}{13.16}+\dfrac{6}{16.19}+\dfrac{6}{19.21}\)

\(\dfrac{1}{2}M=\dfrac{3}{10.13}+\dfrac{3}{13.16}+\dfrac{3}{16.19}+\dfrac{3}{19.21}\)

\(\dfrac{1}{6}M=\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{21}\)

\(\dfrac{1}{6}M=\dfrac{1}{10}-\dfrac{1}{21}\)

\(M=\dfrac{11}{210}:\dfrac{1}{6}=\dfrac{11}{35}\)

\(N=\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{30}\)

\(=\dfrac{1}{20}-\dfrac{1}{30}\)

\(=\dfrac{1}{60}\)

\(\dfrac{M}{N}=\dfrac{11}{35}:\dfrac{1}{60}=\dfrac{132}{7}\)= \(\dfrac{132}{25}\)

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

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

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

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

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

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

Vậy tổng trên nhỏ hơn \(\frac{1}{9}\)

1: 

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

...

\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)

=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)

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

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

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

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

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

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

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

:3 B<1/79 mà 1/79<1/9=>B<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