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18 tháng 2 2017

a ) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{4}{20}+\frac{3}{20}\right)+\left(\frac{8}{21}+\frac{2}{21}-\frac{10}{21}\right)+\left(\frac{2}{5}-\frac{3}{5}\right)\)

\(=\frac{7}{20}+0+\frac{-1}{5}=\frac{7-4}{20}=\frac{3}{20}\)

b ) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{-21}{23}+\frac{-125}{143}\)

\(=0+\frac{-125}{143}=-\frac{125}{143}\)

18 tháng 2 2017

bài 2

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)
=\(1-\frac{1}{2004}=\frac{2003}{2004}\)

19 tháng 2 2018

a) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(=\frac{0}{5}+\frac{0}{21}+\frac{3}{20}\)

\(=0+0+\frac{3}{20}\)

\(=\frac{3}{20}\)

 b) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\frac{125}{93}+\frac{-125}{143}\)

\(=0+\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{17875}{13299}+\frac{-11625}{13299}\)

\(=\frac{6250}{13299}\)

(Hình như câu b hơi sai)

7 tháng 6 2020

\(a)\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{10}{21}+\frac{3}{20}\)

\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(=0+0+\frac{3}{20}\)

\(=\frac{3}{20}\)

26 tháng 3 2018

a)1/1x2+1/2x3+....+1/2003x2004

=1-1/2+1/2-1/3+...+1/2003+1/2004

=1-1/2004

=2004/2004-1/2004

=2003/2004

b)1/1x3+1/3x5+...+1/2003x2005

=1-1/3+1/3-1/5+....+1/2003+1/2005

=1-1/2005

=2005/2005-1/2005

=2004/2005

26 tháng 1 2019

2004/2005

6 tháng 3 2016

Ta có:

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}=\frac{2003}{2004}\)

b,

\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\right).\frac{1}{2}\)

\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\right).\frac{1}{2}\)

\(=\left(1-\frac{1}{2005}\right).\frac{1}{2}=\frac{2004}{2005}.\frac{1}{2}=\frac{1002}{2005}\)

Nhớ nha bạn

3 tháng 3 2020

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2003\cdot2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}=\frac{2003}{2004}\)

b) Đặt A=\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2003\cdot2005}\)

\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{2}{2003\cdot2005}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(2A=1-\frac{1}{2005}\)

\(2A=\frac{2004}{2005}\)

\(A=\frac{2004}{2005}:2=\frac{2004}{2005}\cdot\frac{1}{2}=\frac{1002}{2005}\)

3 tháng 3 2020

a)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=\frac{1}{1}-\frac{1}{2004}\)

\(\Rightarrow=\frac{2003}{2004}\)

b)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003+2005}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(=\frac{1}{1}-\frac{1}{2005}\)

\(\Rightarrow=\frac{2004}{2005}\)

20 tháng 3 2016

giúp nhanh với 

2 tháng 2 2022

\(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\Leftrightarrow\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(\Rightarrow\left(\frac{21}{23}+\frac{-21}{23}\right)+\left(\frac{125}{93}+\frac{-125}{143}\right)=\frac{17000}{3999}\)