Lời giải:

$\frac{A}{2}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$
$=\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}+\frac{9-8}{8\times 9}+\frac{10-9}{9\times 10}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}$

$=1-\frac{1}{9}=\frac{8}{9}$

$\Rightarrow A=2\times \frac{8}{9}=\frac{16}{9}$

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

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

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

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

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

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

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

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

Ta có:

\(\dfrac{1}{3}\times\dfrac{12}{12}=\dfrac{12}{36};\)

\(\dfrac{1}{6}\times\dfrac{6}{6}=\dfrac{6}{36};\)

\(\dfrac{1}{10}\times\dfrac{3}{3}=\dfrac{3}{30};\)

\(\dfrac{1}{15}\times\dfrac{2}{2}=\dfrac{2}{30};\)

\(\dfrac{1}{21}\times\dfrac{4}{4}=\dfrac{4}{84};\)

\(\dfrac{1}{28}\times\dfrac{3}{3}=\dfrac{3}{84};\)

\(A=\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{3}{30}+\dfrac{2}{30}+\dfrac{4}{84}+\dfrac{3}{84}+\dfrac{1}{36}\)

    \(=\left(\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{1}{36}\right)+\left(\dfrac{3}{30}+\dfrac{2}{30}\right)+\left(\dfrac{4}{84}+\dfrac{3}{84}\right)\)

    \(=\dfrac{19}{36}+\dfrac{5}{30}+\dfrac{7}{84}\)

    \(=\dfrac{19}{36}+\dfrac{1}{6}+\dfrac{1}{12}\)

    \(=\dfrac{19}{36}+\dfrac{6}{36}+\dfrac{3}{36}\)

    \(=\dfrac{28}{36}=\dfrac{7}{9}\)

Vậy: \(A=\dfrac{7}{9}\)

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

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

\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=2.\dfrac{2}{5}=\dfrac{4}{5}\)

bn cộng hết lại vs nhau là ra kq!

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)

\(=1-\frac{1}{2020}>1\)

Thank you bạn dcv new ^ ^

nhầm dấu rồi bé hơn 1 chứ:v bạn sửa lại hộ mình nhà

2) A = \(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}\)

=> \(\frac{1}{2}\).A = \(\frac{1}{2}\).\(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}\right)\)

=> \(\frac{1}{2}\).A = \(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

=> \(\frac{1}{2}\).A = \(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

=> \(\frac{1}{2}\).A = \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

=> \(\frac{1}{2}\).A = \(\frac{1}{3}-\frac{1}{9}\)

=> \(\frac{1}{2}\).A = \(\frac{2}{9}\)

=> A = \(\frac{2}{9}:\frac{1}{2}\)

=> A = \(\frac{4}{9}\)

chang hieu cau hoi gi

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

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

A =2\(\times\)( \(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\))

A = 2 \(\times\) ( \(\dfrac{1}{1}\)- \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\))

 A = 2\(\times\)( 1 - \(\dfrac{1}{9}\))

A = 2 \(\times\) \(\dfrac{8}{9}\)

A = \(\dfrac{16}{9}\)

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

\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right).\frac{1}{2}\)

\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)

Trần Thùy Dung: hay cho l.i.k.e

4/5