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A = \(\dfrac{1}{2}\) x 5 + \(\dfrac{1}{5}\) x 8 + \(\dfrac{1}{8}\) x 11 + \(\dfrac{1}{14}\) x 17

A = \(\dfrac{5}{2}\) + \(\dfrac{8}{5}\) + \(\dfrac{11}{8}\) + \(\dfrac{17}{14}\)

A = \(\dfrac{700}{280}\) + \(\dfrac{448}{280}\) + \(\dfrac{385}{280}\) + \(\dfrac{340}{280}\)

\(\Rightarrow\) A = \(\dfrac{1873}{280}\)

A \(=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}\)

A \(=\)\(\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}\right)\)

A \(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\right)\)

A \(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{17}\right)\)

A \(=\dfrac{1}{3}.\dfrac{15}{34}\)

A \(=\dfrac{5}{34}\)

Đặt \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\)

\(A=\dfrac{3}{2}-\dfrac{3}{5}+\dfrac{3}{5}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{14}\)

\(A=\dfrac{3}{2}-\dfrac{3}{14}\)

\(A=\dfrac{21}{14}-\dfrac{3}{14}\)

\(A=\dfrac{18}{14}\)

\(A=\dfrac{9}{7}\)

\(A=1\dfrac{2}{7}\)

\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}\\ =\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\\ =\dfrac{1}{2}-\dfrac{1}{14}\\ =\dfrac{3}{7}\)

Đặt  A = \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{605.608}\)

\(\Rightarrow3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{605.608}\)

\(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{605}-\frac{1}{608}\)

\(3A=\frac{1}{5}-\frac{1}{608}\)

\(A=\left(\frac{1}{5}-\frac{1}{608}\right).\frac{1}{3}=\frac{201}{3040}\)

\(3\times\left(\frac{1}{5\times8}+\frac{1}{8\times11}+....+\frac{1}{97\times100}+x\right)=\frac{319}{100}\)

\(\Rightarrow\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{97\times100}\right)+3\times x=\frac{319}{100}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}+3\times x=\frac{319}{100}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{100}+3\times x=\frac{319}{100}\)

\(\Rightarrow\frac{19}{100}+3\times x=\frac{319}{100}\)

\(\Rightarrow3\times x=\frac{319}{100}-\frac{19}{100}\)

\(\Rightarrow3\times x=3\)

\(\Rightarrow x=3:3\)

\(\Rightarrow x=1\)

Vậy x = 1

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{27}{480}\)

\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{27}{480}.\frac{1}{3}\)

\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{3}{160}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{160}\)

\(\Rightarrow\frac{1}{x+3}=\frac{31}{160}\)

\(\Rightarrow160=31x+93\)

\(\Rightarrow31x=67\)

\(\Rightarrow x=\frac{67}{31}\)

\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+....+\frac{1}{97\cdot100}\)

\(=\frac{5-2}{2\cdot5}+\frac{8-5}{5\cdot8}+\frac{11-8}{8\cdot11}+...+\frac{100-97}{97\cdot100}\)

\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\cdot\frac{49}{100}=\frac{49}{300}\)

\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(=\frac{1}{5}-\frac{1}{2009}\)

\(=\frac{2004}{10045}\)

Đề: Tính

\(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)

Vậy \(A=\frac{2004}{10045}.\)

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)

\(A=\frac{1}{3}\cdot\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

\(A=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\cdot\frac{24}{49}\)

\(A=\frac{8}{49}\)

Đề sai rồi bạn nhé, đề là như thế này:

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)

\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}.\frac{24}{49}\)

\(A=\frac{24}{147}=\frac{8}{49}\)