\(E=\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{6}-1\right)\cdot\left(\dfrac{1}{10}-1\right)\cdot...\cdot\left(\dfrac{1}{45}-1\right)\)

\(E=1\cdot\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{45}\right)\)

\(E=1\cdot\left(\dfrac{1}{45}-\dfrac{1}{3}\right)\div2+1\)

\(E=\dfrac{1}{22}\)

Câu 2:

\(D=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)

Câu 3: 

\(E=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{205}-\dfrac{1}{207}\right)\)

\(=2\cdot\left(1-\dfrac{1}{207}\right)=2\cdot\dfrac{206}{207}=\dfrac{412}{207}\)

Câu 5: 

\(G=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{16}{17}=\dfrac{4}{17}\)

215

215

Cái này

\(S=1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{45}\)

\(S=2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)

\(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

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

\(S=2.\left(1-\frac{1}{10}\right)\)

\(S=2.\frac{9}{10}\)

\(S=\frac{9}{5}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)

= \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

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

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

= \(2.\left(\frac{1}{2}-\frac{1}{10}\right)\)

= \(2.\frac{2}{5}\)

= \(\frac{4}{5}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\\ =\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\\ =2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

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

Lời giải:

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

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

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

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

\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{36}\) + \(\dfrac{1}{45}\)

= \(\dfrac{2}{4}\) + \(\dfrac{2}{6}\) + \(\dfrac{2}{12}\) + \(\dfrac{2}{20}\) + \(\dfrac{2}{30}\) + ... + \(\dfrac{2}{72}\) + \(\dfrac{2}{90}\)

= \(\dfrac{2}{2.2}\) + \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + \(\dfrac{2}{5.6}\) + ... + \(\dfrac{2}{8.9}\) + \(\dfrac{2}{9.10}\)

= 2 (\(\dfrac{1}{2.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + ... + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))