\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=\dfrac{3}{4}\)

\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{10}\right)=\dfrac{3}{4}\Leftrightarrow\dfrac{9}{10}\left(x-\dfrac{1}{3}\right)=\dfrac{3}{8}\)

\(\Leftrightarrow x-\dfrac{1}{3}=\dfrac{5}{12}\Leftrightarrow x=\dfrac{5}{12}+\dfrac{1}{3}=\dfrac{9}{12}=\dfrac{3}{4}\)

Đây là dạng tính nhanh tổng các phân số, trong đó mỗi phân số của tổng có tử số bằng hiệu hai thừa số dưới mẫu và mẫu thứ hai của thừa số này là mẫu số thứ nhất của phân số liền kề với nó. Em tách từng phân số thành hiệu hai phân số mà tử số là 1 còn mẫu số là mẫu hai mẫu số của phân số ban đầu. Triệt tiêu các hạng tử giống nhau ta được tổng cần tìm  

       Dưới đây là cách giải chi tiết em tham khảo nhé em.

A = \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\)+ .....+ \(\dfrac{1}{99\times100}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) +.....+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)

A =  \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)

A = \(\dfrac{99}{100}\)

HD: \(\dfrac{1}{nx\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)

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

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
#kễnh

\(B=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{199\times200}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=1-\dfrac{1}{200}=\dfrac{199}{200}\)

a, \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{5}{12}+\dfrac{19}{30}\)

\(=\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{5}{12}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{19}{30}\right)\)

\(=1+1=2\)

Chúc bạn học tốt!!!

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{1998.1999}+\dfrac{1}{1999.2000}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1998}-\dfrac{1}{1999}+\dfrac{1}{1999}-\dfrac{1}{2000}\)

\(=1-\dfrac{1}{2000}=\dfrac{1999}{2000}.\)

\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{5}{12}+\dfrac{19}{30}\)

\(=\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{5}{12}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{19}{30}\right)\)

\(=\dfrac{12}{12}+\dfrac{30}{30}=1+1=2\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{1998.1999}+\dfrac{1}{1999.2000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{1998}-\dfrac{1}{1999}+\dfrac{1}{1999}+\dfrac{1}{2000}\)\(=1-\dfrac{1}{2000}=\dfrac{1999}{2000}\)

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

=1-1/10

=9/10

\(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{19\cdot20}\)

\(=2\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(=2\cdot\left(1-\dfrac{1}{20}\right)\)

\(=2\cdot\dfrac{19}{20}\)

\(=\dfrac{19}{10}\)

A, \(\left(\dfrac{8}{15}+\dfrac{14}{23}\right)-\left(\dfrac{5}{15}-\dfrac{9}{23}\right)\)

\(=\dfrac{8}{15}+\dfrac{14}{23}-\dfrac{5}{15}+\dfrac{9}{23}\)

\(=\left(\dfrac{8}{15}-\dfrac{5}{15}\right)+\left(\dfrac{14}{23}+\dfrac{9}{23}\right)\)

\(=\dfrac{3}{15}+1\)

\(=1\dfrac{1}{5}\)

B, \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

\(=1-\dfrac{1}{6}\)

\(=\dfrac{5}{6}\)

a) \(=\dfrac{8}{15}+\dfrac{14}{23}-\dfrac{5}{15}+\dfrac{9}{23}\)

\(=\dfrac{8}{15}-\dfrac{5}{15}+\dfrac{14}{23}+\dfrac{9}{23}\)

\(=\dfrac{1}{5}+1\)

\(=\dfrac{6}{5}\)

b)

b) \(=\dfrac{1}{2x}\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{15}\right)\)

\(=\dfrac{1}{2x}\left(2+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}\right)\)

\(=\dfrac{1}{2x}[2\left(1+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)]\)

\(=\dfrac{1}{2x}[2\left(1+\dfrac{1}{3}\right)]\)

\(=\dfrac{1}{2x}\left(2.\dfrac{4}{3}\right)\)

\(=\dfrac{4}{3x}\)