đặt \(M=\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}\)

ta có:

\(M=\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}\)

\(\Leftrightarrow M=\dfrac{3+4}{3.4}-\dfrac{4+5}{4.5}+\dfrac{5+6}{5.6}-\dfrac{6+7}{6.7}+\dfrac{7+8}{7.8}-\dfrac{8+9}{8.9}+\dfrac{9+10}{9.10}\) \(\Leftrightarrow M=\dfrac{3}{3.4}+\dfrac{4}{3.4}-\dfrac{4}{4.5}-\dfrac{5}{4.5}+\dfrac{5}{5.6}+\dfrac{6}{5.6}-\dfrac{6}{6.7}-\dfrac{7}{6.7}+\dfrac{7}{7.8}+\dfrac{8}{7.8}-\dfrac{8}{8.9}-\dfrac{9}{8.9}+\dfrac{9}{9.10}+\dfrac{10}{9.10}\) \(\Rightarrow M=\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{8}+\dfrac{1}{10}+\dfrac{1}{9}\) \(\Rightarrow M=\dfrac{1}{3}+\dfrac{1}{10}\)

\(\Rightarrow M=\dfrac{10}{30}+\dfrac{3}{30}\)

\(\Rightarrow M=\dfrac{13}{30}\)

vậy M = \(\dfrac{13}{30}\)

vậy \(\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}=\dfrac{13}{30}\)

\(\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}=\dfrac{3+4}{3.4}-\dfrac{4+5}{4.5}+\dfrac{5+6}{5.6}-\dfrac{6+7}{6.7}+\dfrac{7+8}{7.8}-\dfrac{8+9}{8.9}+\dfrac{9+10}{9.10}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{7}{30}\)

B = \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\) - \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\)

B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))

B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))

B =  \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{10}\)) 

B = \(\dfrac{1}{12}\) - \(\dfrac{3}{20}\) 

B = - \(\dfrac{1}{15}\)

C = 1/(9.10) - 1/(8.9) - 1/(7.8) - ... - 1/(2.3) - 1/(1.2)

= 1/9 - 1/10 - 1/8 + 1/9 - 1/7 + 1/8 - ... - 1/2 + 1/3 - 1 + 1/2

= 1/9 - 1/10 + 1/9 - 1

= 2/9 - 11/10

= -79/90

chuyện gì ?

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

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

\(=\dfrac{1}{2}-\dfrac{1}{10}\)

\(=\dfrac{2}{5}\)

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

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

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{10}\)

\(\Rightarrow\dfrac{2}{5}\)

3.(1/4.5+1/5.6+...+1/9.10).x=9/2

3.(1/4-1/5+1/5-1/6+...+1/9-1/10).x=9/2

3.(1/4-1/10).x=9/2

3.3/20.x=9/2

9/20.x=9/2

x=9/2:9/20

x=10

Ta có:

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

`@` `\text {Ans}`

`\downarrow`

`a)`

\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)

`=`\(\dfrac{2}{9}\)

Vậy, \(A=\dfrac{2}{9}\)

`b)`

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)

Vậy, \(B=\dfrac{4}{25}\)

`c)`

\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)

`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy, \(C=\dfrac{99}{100}\)