A= \(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+ ... + \(\frac{1}{19.20.21}\)< \(\frac{1}{4}\)

  = 1 - \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{2}\)-  \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\)

  = 1 - ( \(\frac{1}{2}-\frac{1}{3}\)+ \(\frac{1}{2}-\frac{1}{3}\)) + ... + ( \(\frac{1}{19}-\frac{1}{20}+\frac{1}{19}-\frac{1}{20}\))  - \(\frac{1}{21}\)

  = 1 - \(\frac{1}{21}\)

  =  \(\frac{20}{21}\)<  \(\frac{1}{4}\)

=> Đề bài có sai ko bạn?

\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right).x=5\)

\(\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{21-19}{19.20.21}\right).x=5\)

 \(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right).x=5\)

 \(\left(\frac{1}{1.2}-\frac{1}{20.21}\right).x=5\)

 \(\frac{209}{420}.x=5\)

\(\Rightarrow x=5\div\frac{209}{420}=\frac{2100}{209}\)

\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right).x=5\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}\right).2.x=5\)

\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{19.20}-\frac{1}{20.21}\right)\right).x.2=5\)

\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\right).x=5\div2\)

\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{20.21}\right)\right).x=2,5\)

\(\left(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{420}\right)\right).x=2,5\)

\(\left(\frac{1}{2}\times\frac{209}{420}\right)\times x=2,5\)

\(\frac{209}{840}\times x=2,5\)

\(x=2,5\div\frac{209}{840}=10\frac{10}{209}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+....+\frac{1}{98.99.100}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\)

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

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

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(=\frac{1}{2}.\frac{4949}{9900}\)

\(=\frac{1}{19800}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(=\frac{1}{2}.\frac{4949}{9900}\)

\(=\frac{4949}{19800}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2004.2005.2006}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2004.2005}-\frac{1}{2005.2006}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2005.2006}\right)\)

\(=\frac{1}{4}-\frac{1}{2.2005.2006}\)

\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{2018\cdot2019\cdot2020}\)

\(=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2018\cdot2019\cdot2020}\right]\)

\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right]\)

Đến đây tự tính được rồi:v

   Đặt tổng trên là A

Ta có:

\(2A=2\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2018\cdot2019\cdot2020}\right)\)

\(=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2018\cdot2019\cdot2020}\)

\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\)

\(=\frac{1}{2}-\frac{1}{2019\cdot2020}\)

\(A=\left(\frac{1}{2}-\frac{1}{2019\cdot2020}\right)\div2\)

        *Làm tiếp*

                                          \(#Louis\)

Ta có công thức:

\(\frac{a}{c.\left[c+1\right].\left[c+2\right]}=\frac{a}{2}\left[\frac{1}{c.\left[c+1\right]}-\frac{1}{\left[c+1\right].\left[c+2\right]}\right]\)

vậy

\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{11.12}-\frac{1}{12.13}\right]\)

\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{12.13}\right]\)

\(C=\frac{1}{2}.\frac{77}{156}=\frac{77}{312}\)

mình làm đầu tiên đó, 

Chúc bạn học tốt !

\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{11.12.13}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{11.12}-\frac{1}{12.13}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{156}\right)\)

\(=\frac{1}{2}\cdot\frac{77}{156}\)

\(=\frac{77}{312}\)

Mình không chép đề bài nhé :
Gọi biểu thức là A :
Ta có : 2A=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
=\(\frac{1}{1.2}-\frac{1}{49.50}\)( Rút gọn đi ta được cái này )
=1/2 - 1/2450
Vậy A = (1/2 - 1/2450):2
 

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)

\(=\frac{1}{2}.\left(\frac{741}{1482}-\frac{1}{1482}\right)\)

\(=\frac{1}{2}.\frac{740}{1482}\)

\(=\frac{185}{741}\)

Chúc bạn học tốt !!! 

Đặt 1/1.2.3 + 1/2.3.4 + ...+ 1/37.38.39 = A

Ta có : 2A = 2/1.2.3 + 2/2.3.4 +...+ 2/37.38.39

         2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/37.38 - 1/38.39

         2A = 1/1.2 - 1/38.39

         2A = 740/1482 = 370/741

           A= 370/741 . 1/2 =........