M = \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2015.2017}\)4/1.3 + 4/3.5 + 4/5.7 + ... + 4/2015.2017

M = \(2.\frac{2}{1.3}+2.\frac{2}{3.5}+2.\frac{2}{5.7}+...+2.\frac{2}{2015.2017}\) 2 . 2/1.3 + 2 . 2/3.5 + 2 . 2/5.7 + ... + 2 . 2/2015.2017

M = 2 . ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2015.2017 )

M = 2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2015 - 1/2017 )

M = 2 . ( 1 - 1/2017 )

M = 2 . 2016/2017

M = 4032/2017

\(M=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(M=2\left(1-\frac{1}{2017}\right)\)

\(M=\frac{4032}{2017}\)

\(M=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2015.2017}\)

\(M=4\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\right)\)

\(M=4.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015.2017}\right)\)

\(M=2\left(1-\frac{1}{2017}\right)\)

\(M=2.\frac{2016}{2017}\)

\(M=\frac{4032}{2017}\)

A=4/3+9/8+16/15+..............+4064256/4064255

A=1+1/3+1+1/8+1/15+...............+1/4064255

A=(1+1+...+1)+(1/3+1/8+...+1/406255)          (có 2015 số 1)

A=2015+(1/1.3+1/2.4+...........+1/2015.2017)
A=2015+1/2(1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+....+1/2012-1/2014+1/2013-1/2015+1/2014-1/2016+1/2015-1/2017)

A=2015+1/2(1+1/2-1/2016-1/2017)

A=2015,749504

                                k cho mình nhé mình k lại cho

ai làm đúng tích luôn

a=1008/2017

Đặt A = \(\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2010.2012+2}{2011^2}+\frac{2015.2017+2}{2016^2}\)

\(=\frac{\left(2-1\right)\left(2+1\right)+2}{2^2}+\frac{\left(3-1\right)\left(3+1\right)}{3^2}+...+\frac{\left(2016-1\right)\left(2016+1\right)+2}{2016^2}\)

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

\(=\frac{2^2+1}{2^2}+\frac{3^2+1}{3^2}+...+\frac{2016^2+1}{2016^2}\)

\(=\left(1+1+...+1\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}\right)\)(2015 hạng tử 1)

\(=2015+\left(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2016.2016}\right)\)

 \(< 2015+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\right)\)

\(=2015+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)=2015+\left(1-\frac{1}{2016}\right)\)

= 2015 + 1 + 1/2016

= 2016 + 1/2016 < 2017 

=> A < 2017 (ĐPCM)

\(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{2015.2017}\)

\(=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2015.2017}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{2017}\right)\)

\(=\frac{3}{2}.\frac{2016}{2017}\)

\(=\frac{3024}{2017}\)

_Chúc bạn học tốt_

(X-21.23):11+39=40

\(\frac{3}{1.3}\)+  \(\frac{3}{3.5}\)+ ... + \(\frac{3}{2015.2017}\) = (3 - 1) : 2 + (1 - \(\frac{3}{5}\)) : 2 + ... + (\(\frac{3}{2015}\)- \(\frac{3}{2017}\)) : 2

                                                                   = (3 - 1 + 1 -\(\frac{3}{5}\)+ ... + \(\frac{3}{2015}\) - \(\frac{3}{2017}\)) : 2

                                                                   = (3 - \(\frac{3}{2017}\)) : 2 

                                                                   = \(\frac{6048}{2017}\) : 2

                                                                   = \(\frac{3024}{2017}\)

=1-1/3+1/3-1/5+...+1/2015-1/2017

=1-1/2017

=2016/2017

http://olm.vn/hoi-dap/question/161872.html

B=1/1.3+1/3.5+...+1/2015.2017

B= 1/2 . 2. ( 1/1.3+1/3.5 + .... + 1/2015 .2017)

B = 1/2 . ( 2/1.3 + 2/3.5 + ......+ 2/2015.2017)

B = 1/2. ( 1/1+ -1/3 + 1/3 + -1/5 + 1/5 +....+ 1/2015 + -1/2017)

B= 1/2 . ( 1/1 + -1 / 2017) = 1/2 . 2016 / 2017 = 2016 / 4034

Vậy B = 2016 / 4034 nha bn.pham hong thai