\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)

\(\frac{A}{2}=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)

\(\frac{A}{2}=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{11\cdot12}\)

\(\frac{A}{2}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)

\(\frac{A}{2}=\frac{1}{4}-\frac{1}{12}\)

\(\Rightarrow A=\frac{2}{4}-\frac{2}{12}=\frac{16}{48}\)

\(B=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{55}\)

\(\frac{B}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\)

\(\frac{B}{2}=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\)

\(\frac{B}{2}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(\frac{B}{2}=\frac{1}{3}-\frac{1}{11}\)

\(\Rightarrow B=\frac{2}{3}-\frac{2}{11}=\frac{16}{33}\)

Mà \(\frac{16}{48}< \frac{16}{33}\Rightarrow A< B\)

Vậy : A < B

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\)

\(A=2\times\dfrac{1}{2}\times\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\right)\)

\(A=2\times\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\right)\)

\(A=2\times\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}\right)\)

\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)

\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{11}\right)\)

\(A=2\times\dfrac{9}{22}\)

\(A=\dfrac{9}{11}\)

\(A=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{55}\)  => \(\frac{A}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)

=> \(\frac{A}{2}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}=\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)

=> \(A=\frac{16}{33}\)

do những số đó bé hơn 1 nên cộng lại vẫn bé hơn 1

  A =  \(\dfrac{1}{3}\) +    \(\dfrac{1}{6}\) +  \(\dfrac{1}{10}\)  + \(\dfrac{1}{15}\) + ..+ \(\dfrac{1}{55}\)+ \(\dfrac{1}{66}\)

A  = 2  \(\times\) ( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\)  + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) +...+ \(\dfrac{1}{110}\) + \(\dfrac{1}{132}\))

A  = 2 \(\times\) ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +  \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\) +...+ \(\dfrac{1}{10.11}\)+ \(\dfrac{1}{11.12}\))

A = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +...+ \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)+ \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\))

A = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{12}\))

A = 1 - \(\dfrac{1}{6}\) < 1

Vậy A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ...+ \(\dfrac{1}{55}\)+ \(\dfrac{1}{66}\) < 1 

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{55}\)

\(2+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{110}\)

\(2+2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{55}\right)\)

\(2+2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\right)\)

\(2+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(2+2.\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(2+2.\frac{9}{22}\)

\(2+\frac{9}{11}\)

\(\frac{31}{11}\)

Gọi biểu thức trên là A

Ta có :

\(A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{55}\)

\(\frac{1}{2}A=\frac{1}{2}\times\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}\right)\)

\(\frac{1}{2}A=\frac{1}{2}\times1+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{2}\times\frac{1}{55}\)

\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)

\(\frac{1}{2}A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{10\times11}\)

\(\frac{1}{2}A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(\frac{1}{2}A=1-\frac{1}{11}\)

\(\frac{1}{2}A=\frac{10}{11}\)

\(A=\frac{10}{11}\div\frac{1}{2}\)

\(A=\frac{20}{11}\)

Ta đặt:

\(A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{55}\)

\(\Rightarrow\frac{1}{2}A=1.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{6}+\frac{1}{2}.\frac{1}{10}+\frac{1}{2}.\frac{1}{15}+...+\frac{1}{2}.\frac{1}{55}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{1}-\frac{1}{11}=\frac{10}{11}\Leftrightarrow A=\frac{10}{11}:\frac{1}{2}=\frac{20}{11}\)

\(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+....+\frac{1}{55}\)

=\(\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+....+\frac{2}{110}\)

=\(2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{10.11}\right)\)

=\(2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+..+\frac{1}{10}-\frac{1}{11}\right)\)

= \(2.\left(\frac{1}{3}-\frac{1}{11}\right)\)

=\(2.\frac{8}{33}\)

= \(\frac{16}{33}\)

Bài làm:

Ta có: \(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{66}\)

\(=\frac{1}{1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{11.6}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.1.3}+\frac{1}{2.3.2}+...+\frac{1}{2.11.6}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{12}\right)\)

\(=\frac{1}{2}.\frac{11}{12}\)

\(=\frac{11}{24}\)

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)

\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)

\(=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{10\times11}+\frac{1}{11\times12}\right)\)

\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)

\(=2\times\left(1-\frac{1}{12}\right)\)

\(=2\times\frac{11}{12}\)

\(=\frac{11}{6}\)

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)

\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)

\(=2\left(1-\frac{1}{12}\right)=2.\frac{11}{12}=\frac{22}{12}=\frac{11}{6}\)

A =\(2.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+......+\dfrac{1}{156}\right)\)

A =\(2.\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+..........+\dfrac{1}{12.13}\right)\)

A =2.\(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)

A=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)

tớ làm hơi gọn nên có gì kho hiểu thì nói tớ