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 = \(\dfrac{16}{9}\)

Lời giải:

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

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

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

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

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                                                 

giúp em điiiiiii

thì cộng vào nó vẫn ra 3/10 mà

Kết quả là: 9/5 bn nhé!

Chúc bn học tốt!

1+1/3+1/6+1/10+1/15+1/21+1/28+1/38+1/45=9/5

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

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

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

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

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

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

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

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

\(A=\left(\frac{1}{3}-1\right)\left(\frac{1}{6}-1\right)\left(\frac{1}{10}-1\right)....\left(\frac{1}{36}-1\right)\)

\(=\frac{-2}{3}.\frac{-5}{6}.\frac{-9}{10}...\frac{-35}{36}\)

\(=-\left(\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{70}{72}\right)=-\left(\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{7.10}{8.9}\right)\)

\(=-\left(\frac{1.4.2.5.3.6...7.10}{2.3.3.4.4.5...8.9}\right)=-\frac{\left(1.2.3.4..7\right)\left(4.5.6...10\right)}{\left(2.3.4...8\right)\left(3.4.5...9\right)}=-\frac{1.10}{8.3}=\frac{-10}{24}\)

Đặt biểu thức là A .

Ta có :

\(A=\left(\frac{1}{3}-1\right)\left(\frac{1}{6}-1\right)\left(\frac{1}{10}-1\right)...\left(\frac{1}{36}-1\right)\)

\(A=\frac{-2}{3}\cdot\frac{-5}{6}\cdot\frac{-9}{10}\cdot....\cdot\frac{-36}{36}\)

\(A=-\left(\frac{4}{6}\cdot\frac{10}{12}\cdot\frac{18}{20}\cdot...\cdot\frac{70}{72}\right)\)

\(A=-\left(\frac{1.4}{2.3}\cdot\frac{2.5}{3.4}\cdot\frac{3.6}{4.5}...\frac{7.10}{8.9}\right)\)

\(A=-\left(\frac{1.4.2.5.3.6...7.10}{2.3.3.4.4.5...8.9}\right)\)

\(A=-\frac{\left(1.2.3.4....7\right)\left(4.5.6....10\right)}{\left(2.3.4....8\right)\left(3.4.5....9\right)}\)

\(A=-\frac{1.10}{8.3}\)

\(A=\frac{-10}{24}\)