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21 tháng 7 2021

`A=1/(1.3)+1/(3.5)+....+1/(2009.2011)`

`=> 2A=2/(1.3)+1/(3.5)+...+2/(2009.2011)`

`=1-1/3+1/3+1/5+.....+1/2009-1/2011`

`=1-1/2011`

`=2010/2011`

`=> A=1005/2011`

21 tháng 7 2021

Dựa trên công thức: \(\dfrac{a}{n.\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\), ta có:

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}\)

\(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2009.2011}\right)\)

\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

=\(\dfrac{1}{2}.\left(1-\dfrac{1}{2011}\right)\)

\(\dfrac{1}{2}.\left(\dfrac{2010}{2011}\right)\)

\(\dfrac{2010}{4022}\)

= tự rút gọn nhé

Hok tốt!

23 tháng 8 2023

\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)

24 tháng 8 2023

Thanks

16 tháng 2 2022

1100444-88888=

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)

\(\frac{10}{22}\)

23 tháng 7 2017

a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=5\left(1-\dfrac{1}{100}\right)\)

\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)

b, \(C=1.2.3+2.3.4+...+8.9.10\)

\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)

\(C=\dfrac{8.9.10.11}{4}=1980.\)

c, https://hoc24.vn/hoi-dap/question/384591.html

Câu này bạn vào đây mình đã giải câu tương tự nhé.

23 tháng 7 2017

\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)

\(\Leftrightarrow A=\dfrac{99}{20}\)

2 tháng 8 2018

\(B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+....+\dfrac{2}{99.101}\)

\(B=2.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)

\(B=\dfrac{2}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}\right)\)

\(B=1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(B=1.\left(1-\dfrac{1}{101}\right)\)

\(B=1.\dfrac{100}{101}\)

\(B=\dfrac{100}{101}\)

cảm mơn bn

bn giúp mình pB nha

5 tháng 5 2022

`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`

`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`

`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`

`=(1-1/7).[130-4-125]/3`

`=6/7 . 1/3 = 2/7`

____________________________________________________

`8/9+1/9 . 2/9+1/9 . 7/9`

`=8/9+1/9.(2/9+7/9)`

`=8/9+1/9 . 9/9`

`=8/9+1/9=9/9=1`

a) \(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)

\(=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}\)

\(=-\dfrac{5}{24}\)

 

b) \(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)\)

\(=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}\)

\(=\dfrac{6}{7}\cdot\dfrac{1}{3}\)

\(=\dfrac{2}{7}\)

 

\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)

\(=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}\)

\(=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}\)

\(=1\)

31 tháng 8 2015

\(2.A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2007.2009}+\frac{2}{2009.2011}\)

\(2.A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2009-2007}{2007.2009}+\frac{2011-2009}{2009.2011}\)

\(2.A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2011}=1-\frac{1}{2011}=\frac{2010}{2011}\)

=>A = \(\frac{2010}{2011}:2=\frac{1005}{2011}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{13\cdot15}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)

6 tháng 4 2022

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{5}{15}-\dfrac{1}{15}=\dfrac{4}{15}\)