\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(=1-\frac{1}{1000}+1\)

\(=\frac{1000}{1000}-\frac{1}{1000}+\frac{1000}{1000}\)

\(=\frac{1999}{1000}\)

Tham khảo nhé~

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

= \(1-\frac{1}{1000}+1\)

= \(\frac{1999}{1000}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\right)+1\)

\(=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{999}-\frac{1}{1000}\right)+1\)

\(=\left(1-\frac{1}{1000}\right)+1\)

\(=\frac{999}{1000}+1\)

\(=\frac{1999}{1000}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=\frac{1}{2}-\frac{1}{7}\)

\(=\frac{7}{14}-\frac{2}{14}\)

\(=\frac{5}{14}\)

#)Giải :

Gọi các tổng trên là A

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{6}\)

\(\Rightarrow A=\frac{1}{3}\)

         #~Will~be~Pens~#

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)\(=\frac{1}{2}-\frac{1}{7}=\frac{7}{14}-\frac{2}{14}=\frac{5}{14}\)

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

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

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

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

\(=\frac{4}{10}\)

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

ai k tôi k lại

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

1/2.3 + 1/3.4 + 1/4.5 + ... + 1/9.10

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10

= 1/2 - 1/10

= 5/10 - 1/10

= 4/10 = 2/5

=>-(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5)

=>-(1-1/5)

=>-4/5

\(\:\frac{-1}{1.2}+\frac{-1}{2.3}+\frac{-1}{3.4}+\frac{-1}{4.5}\)

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

=\(-1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)

=\(-1\left(1-\frac{1}{5}\right)\)

=\(-1\times\frac{4}{5}\)

=\(\frac{-4}{5}\)

Ta có 

\(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\)

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

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

Vậy \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)\(=\frac{19}{10}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)

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

\(=1-\frac{1}{2017}=\frac{2016}{2017}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

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

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

1/1.2+1/2.3+1/3.4+1/4.5+1/5.6

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6

=1-1/6

=5/6

1/1 - 1/2+1/2 -1/3+1/3 - 1/4+1/4 - 1/5+1/5 - 1/6

=1-( -1/2+1/2+-1/3+1/3+-1/4+1/4+-1/5+1/5)-1/6

=1 - 1/6

=6/6 - 1/6

=5/6

\(x-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{4.5}\)

\(x-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}=0\)

\(x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)=0\)

\(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)=0\)

\(x-\left(1-\frac{1}{5}\right)=0\)

\(x-\frac{4}{5}=0\)

\(x=\frac{4}{5}\)

\(x-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{4.5}\)

\(x-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{4}-\frac{1}{5}\)

\(x-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}=\frac{-1}{5}\)

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

\(x=\frac{4}{5}\)

\(x-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{4.5}\)

\(\Rightarrow x-\text{​​}\text{​​}\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}=0\)

\(\Rightarrow x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)=0\)

\(\Rightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)=0\)

\(\Rightarrow x-\left(1-\frac{1}{5}\right)=0\)

\(\Rightarrow x-\frac{4}{5}=0\)

\(\Rightarrow x=\frac{4}{5}\)

Vậy  \(x=\frac{4}{5}\)

_Chúc bạn học tốt_