\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)

\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)

Lâm đi là: 35 phút +2 giờ 20phút =2 giờ 55 phút

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(A=1-\frac{1}{2017}\)

\(A=\frac{2016}{2017}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2016.2018}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2016.2018}\)

\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2016}-\frac{1}{2018}\)

\(2B=1-\frac{1}{2018}\)

\(B=\frac{2017}{\frac{2018}{2}}\)

a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{2009.2010}\)

A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)

A=1-\(\frac{1}{2010}\)=\(\frac{2009}{2010}\)

c)C=\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+......+\frac{1}{2006.2008}\)

C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2006}-\frac{1}{2008}\))

C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{2008}\))

C=\(\frac{1}{2}\).\(\frac{1003}{2008}\)=\(\frac{1003}{4016}\)

Câu b mình chưa nghĩ ra

Chúc bạn học tốt!

a) A = \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + ...+ \(\frac{1}{2009.2000}\)

= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ... + \(\frac{1}{2009}\) - \(\frac{1}{2000}\)

= 1 - \(\frac{1}{2000}\) = \(\frac{1999}{2000}\)

b) B = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + \(\frac{1}{3.4.5}\) + ... + \(\frac{1}{1998.1999.2000}\)

= \(\frac{1}{2}\) ( \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + \(\frac{2}{3.4.5}\) + ... + \(\frac{2}{1998.1999.2000}\))

= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + \(\frac{1}{3.4}\) - \(\frac{1}{4.5}\) + ... + \(\frac{1}{1998.1999}\) - \(\frac{1}{1999.2000}\))

= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{1999.2000}\))

= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{3998000}\))

= \(\frac{1}{4}\) - \(\frac{1}{7996000}\) = ?

c) C = \(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + \(\frac{1}{6.8}\) + ... + \(\frac{1}{2006.2008}\)

= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{4}\)) + \(\frac{1}{2}\)(\(\frac{1}{4}\) - \(\frac{1}{6}\)) + ... + \(\frac{1}{2}\)(\(\frac{1}{2006}\) - \(\frac{1}{2008}\))

= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\) + ... + \(\frac{1}{2006}\) - \(\frac{1}{2008}\))

= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{2008}\))

= \(\frac{1}{2}\) . \(\frac{1003}{2008}\) = \(\frac{1003}{4016}\).

\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)x=\frac{1}{3}\left(2014.2015.2016-2013.2014.2015........+2.3.4-1.2.3+1.2.3-0.1.2\right)\)

\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)

\(x=\frac{1}{3.2029104}.2014^2.2015^2.2016=\)

\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)

vào câu hỏi tương tự nha bạn

Làm bậy, mà đúng

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ … + \(\frac{1}{99.100}\)

= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ … + \(\frac{1}{99}\)- \(\frac{1}{100}\)

= \(\frac{1}{1}\)- \(\frac{1}{100}\)

= \(\frac{99}{100}\)

    1/1 . 2 + 1/ 2 . 3 + 1/ 3 . 4 + ... + 1/99 . 100

= 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100

= 1/1 - 1/100 

= 100/100 + -1/100

= 99/100

            #Hoq chắc _ Baccanngon