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13 tháng 2 2018

\(B=2016.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)

\(2016.\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{2015^2}{2014.2016}\)

\(2016.\frac{2.3.4....2015}{1.2.3.4.5...2014.2015.2016}.\frac{2.3.4....2015}{3.4.5...2014}\)

\(2016.\frac{1}{2016}.2.2015=2.2015=4030\)

6 tháng 4 2022

\(C=\left(1+\frac{1}{1.3}\right)\)\(.\left(1+\frac{1}{2.4}\right)\)\(.\left(1+\frac{1}{3.5}\right)\)\(.\left(1+\frac{1}{2014.2016}\right)\)

   \(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2015^2}{2014.2016}\)

   \(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)

   \(=\frac{\left(2.3.4...2015\right).\left(2.3.4...2015\right)}{\left(1.2.3...2014\right).\left(3.4.5...2016\right)}\)

   \(=\frac{2015.2}{2016}\)

    \(=...\)(tự tinhs)

12 tháng 3 2016

giúp mới mình sẽ tích

12 tháng 3 2016

bn viết rõ đề đi

10 tháng 1 2016

5435

tích mình nha

10 tháng 1 2016

luu y : dau /la phan cach giua mau so va tu so

Có \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..........\)\(\left(1+\frac{1}{2014.2016}\right)\)

=\(\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)....\left(\frac{2014.2016}{2014.2016}+\frac{1}{2014.2016}\right)\)

=\(\left(\frac{2^2-1}{1.3}+\frac{1}{2.4}\right)\left(\frac{3^2-1}{2.4}+\frac{1}{2.4}\right)......\left(\frac{2015^2-1}{2014.2016}+\frac{1}{2014.2016}\right)\)

=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2015.2015}{2014.2016}\)

=\(\frac{2.2.3.3.....2015.2015}{1.3.2.4....2014.2015}\)

=\(\frac{\left(2.3...2015\right).\left(2.3.....2015\right)}{\left(1.2....2014\right).\left(3.4.....2016\right)}=\frac{2015.2}{2016}=\frac{4030}{2016}\)

27 tháng 2 2017

Chào bạn !Mình kết bạn nha!

27 tháng 2 2017
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                               
\(B=\frac{3+1}{3}+\frac{8+1}{8}+...+\frac{2014.2016+1}{2014.2016}+\frac{2015.2017+1}{2015.2017}\)         
\(=\frac{2^2}{3}+\frac{3^2}{8}+....+\frac{2015^2}{2014.2016}+\frac{2016^2}{2015.2017}\)         
=\(\frac{2.2}{3}+\frac{3.3}{2.4}+...+\frac{2015.2015}{2014.2016}+\frac{2016.2016}{2015.2017}\)         
=\(\frac{\left(2.3....2015.2016\right)+\left(2.3.....2015.2016\right)}{\left(1.2.3.....2014.2015\right)+\left(3.4....2016.2017\right)}\)         
=\(2016+\frac{2}{2017}\)