a,\(\frac{2015.2016+2015-1}{2014+2015.2016}=\frac{2015.2016+2014}{2014+2015.2016}=1\)\(1\)

b,\(=1-\frac{1}{5}+\frac{1}{5}...-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)

c,\(=\frac{12}{35}+\frac{12}{35}+\frac{12}{35}+\frac{12}{35}=\frac{12}{35}.4=\frac{48}{35}\)

48/35 nha

b>=1-1/5+1/5-1/9+...+1/2011-1/2015

 =1-1/2015

=2014/2015

c>=12/35+12/35+12/35+12/35

=12/35x4

=48/35

C=(1+1+1+...+1)+(1/1*3+1/2*4+1/3*5+...+1/2015*2017+1/2015*2017)

C=2015+(2/1*3+2/2*4+2/3*5+...+2/2015*2017+2/2015*2017):2

C=2015+(1-1/3+1/2-1/4+...+1/2015-1/2017+1/2015-1/2017):2

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

cai nay thi ban tu tinh lay 

nho k cho minh voi nhe

Dương Thanh Minh lạc đề rùi

=(1.3+1/1.3).(2.4+1/2.4).....(2015.2017+1/2015.2017)

=(2.2/1.3).(3.3+1/2.4)....(2016.2016/2015.2017)

=(2.3.4.....2016/1.2.3...2015).(2.3.4....2016/3.4.5...2017)

=2016.(2/2017)

=Lấy máy tính bấm nhé

(^_^)

B = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)

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

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

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

B = \(\frac{1}{2}.\frac{2016}{2017}\)

B = \(\frac{1008}{2017}\)

Vậy B = \(\frac{1008}{2017}\)

Chúc bạn học tốt . Có bài gì khó mik sẽ giúp bạn ( Chỉ toán 6 hoặc 7 trở xuống thui đó )

\(B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{2015\cdot2017}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{2015\cdot2017}\right)\)

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

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

\(B=\frac{1}{2}\cdot\frac{2016}{2017}\)

\(B=\frac{1008}{2017}\)

\(B=\frac{1}{1.3}+\frac{1}{3.7}+...+\frac{1}{2005.2007}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2015.2017}\)

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

\(2B=1-\frac{1}{2017}\)

\(2B=\frac{2016}{2017}\)

\(B=\frac{2016}{2017}:2\)

\(B=\frac{1008}{2017}\)

Vậy  \(B=\frac{1008}{2017}\)

A = 

A = \(1-\frac{1}{2018}\)

A = \(\frac{2017}{2018}\)

Có : 

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

2.B = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

2.B = \(1-\frac{1}{2017}\)

2.B = \(\frac{2016}{2017}\)

B = \(\frac{2016}{2017}:2=\frac{1008}{2017}\)

Có :

3.C = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2017.2020}\)

3.C = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\)

3.C = \(\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)

C = \(\frac{2019}{2020}:3=\frac{673}{2020}\)

a=1/1-1/2+1/2-1/3+...+1/2017-1/2018

=1/1-1/2018

=kq

may bai duoi lam tuong tu nha

mình chưa điền kết quả ban tu dien nha 

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(A=\frac{1}{1}-\frac{1}{2018}=\frac{2017}{2018}\)

\(B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2015\cdot2017}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2015\cdot2017}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

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

\(B=\frac{1}{2}\cdot\frac{2016}{2017}\)

\(B=\frac{1008}{2017}\)

\(C=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{2017\cdot2020}\)

\(C=\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{2017\cdot2020}\right)\)

\(C=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\right)\)

\(C=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{2020}\right)\)

\(C=\frac{1}{3}\cdot\frac{2019}{2020}\)

\(C=\frac{673}{2020}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{2016\cdot2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{2016}-\frac{1}{2017}\)

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

= 1/1-1/2+1/2-1/3+1/3-............-1/2017

=1-1/2017

=2016/2017

\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+..+\(\frac{1}{2006.2007}\)

=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+..+\(\frac{1}{2006}\)-\(\frac{1}{2007}\)

=1-\(\frac{1}{2007}\)=\(\frac{2006}{2007}\)

\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}+\frac{10}{9.11}+...+\frac{2016}{2015.2017}\)

\(=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

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

\(=2.\frac{2014}{6051}\)

\(=\frac{4028}{6051}\)

\(\Rightarrow BT>\frac{1}{6}\)

Đặt \(A=1.2.3+2.3.4+3.4.5+...+2015.2016.2017\)

=>\(4A=1.2.3.4+2.3.4.4+3.4.5.4+...+2015.2016.2017.4\)

=>\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)\)

\(+...+2015.2016.2017.\left(2018-2014\right)\)

=>\(4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5\)

\(+...+2015.2016.2017.2018-2014.2015.2016.2017\)

=>\(4A=2015.2016.2017.2018\Rightarrow A=\frac{2015.2016.2017.2018}{4}\)