Bn tham khảo nhé : 

2 / 2 . 3 + 2 /3 . 4 + 2 / 4 .5  + ... + 2 / 2017 . 2018

= 2 . ( 1/2 . 3 + 1/3 . 4 + 1/4 . 5 + ... + 1/ 2017 . 2018 

= 2 . ( 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/2017 - 1/2018 ) 

= 2 . ( 1/2 - 1/2018)

= 2 . 1008/2018

= 2016/2018

= 1008/1009

\(2\times(\frac{1}{2\times3}\times\frac{1}{3\times4}\times...\times\frac{1}{2017\times2018}))\)

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

\(2\times(\frac{1}{2}-\frac{1}{2018})\)

\(2\times\frac{504}{1009}=\frac{1008}{1009}\)

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.......+\frac{2}{2017.2018}\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{2017.2018}\right)\)

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

\(=2\left(\frac{1}{2}-\frac{1}{2018}\right)\)

\(=2\left(\frac{1009}{2018}-\frac{1}{2018}\right)\)

\(=2.\frac{1008}{2018}\)

\(=\frac{1008}{1009}\)

cho bài kham khảo nè :

A=1.2+2.3+3.4+4.5+...+2017.2018
=> 3A=1.2.3+2.3.3+3.4.3+4.5.3+...+2017.2018.3
3A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+...+2017.2018.(2019-2016)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+2017.2018.2019-2016.2017.2018
3A=(1.2.3+2.3.4+3.4.5+4.5.6+...+2017.2018.2019)-(1.2.3+2.3.4+3.4.5+...+2016.2017.2018)
=> 3A=2017.2018.2019 => \(A=\frac{2017.2018.2019}{3};B=\frac{2018^3}{3}=\frac{2018.2018.2018}{3}\)

Ta có: 2017.2019=2017(2018-1)=2017.2018+2017<2017.2018+2018=2018(2017+1)=2018.2018
=> 2017.2018.2019<2018.2018.2018
=> A<B

thank nha

A=1.2+2.3+3.4+...+2017.2018

3A=1.2.3+2.3.3+3.4.3+...+2017.2018.3

3A=1.2.3+2.3.(4−1)+3.4.(5−2)+...+2017.2018.(2019−2016)

3A=1.2.3+2.3.4−1.2.3+3.4.5−2.3.4+...+2017.2018.2019−2016.2017.2018

⇒3A=2017.2018.2019

⇒A=2017.2018.20193

A=2017.2018.20193;B=201833=2018.2018.20183

A=2739315938;B=2739316611

⇒A<B

\(A=1.2+2.3+3.4+4.5+............+2017.2018\)

\(3A = 1.2.3 + 2.3.4 +..............+ 2017.1018.3\)

\(3A = 1.2.3 + 2.3.(4-1) + .............. + 2017.2018.(2019-2016)\)

\(3A = 1.2.3 + 2.3.4 - 1.2.3 + ............. + 2017.2018.2019 - 2016.2017.2018\)

\(3A = 2017.2018.2019\)

\(A = \frac{2017.2018.2019}{3}\)

\(B =\frac {2018^3}{3}\)

đến đây ko bt lm

\(S=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

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

\(\Rightarrow S=\frac{1}{2}-\frac{1}{2018}\)

\(\Rightarrow S=\frac{1008}{2018}\)

bạn rút gọn nốt nha mk ko có máy tính

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

\(S=\frac{1}{2}-\frac{1}{2018}\)

\(S=\frac{504}{1009}\)

HK TỐT NHÉ

S = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\)+ ..... + \(\frac{1}{2017.1018}\)

S = \(\frac{1}{2}\) - \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{5}\) + .....+ \(\frac{1}{2017}\)- \(\frac{1}{2018}\)

S = \(\frac{1}{2}\) - \(\frac{1}{2018}\)

S = \(\frac{1008}{2018}\)

CHÚC BẠN HỌC GIỎI

A=1.2+2.3+3.4+4.5+...+2017.2018

=> 3A=1.2.3+2.3.3+3.4.3+4.5.3+...+2017.2018.3

3A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+...+2017.2018.(2019-2016)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+2017.2018.2019-2016.2017.2018

3A=(1.2.3+2.3.4+3.4.5+4.5.6+...+2017.2018.2019)-(1.2.3+2.3.4+3.4.5+...+2016.2017.2018)

=> 3A=2017.2018.2019  => \(A=\frac{2017.2018.2019}{3}\);  \(B=\frac{2018^3}{3}=\frac{2018.2018.2018}{3}\)

Ta có: 2017.2019=2017(2018-1)=2017.2018+2017<2017.2018+2018=2018(2017+1)=2018.2018

=> 2017.2018.2019<2018.2018.2018

=> A<B

Bui The Hao lam dung roi

mk cung dang can bai nay

Thanks vi da dang honganh

Ta có : A=1.2+2.3+3.4+....+2015.2016

=>3A= 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + ... + 2017.2018.3

=>3A= 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5-2 ) + 4.5.( 6-3 ) + ... 2017 . 2018 . ( 2019 - 2016 )

=>3A=-1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 + 4.5.6 - 4.5.3 +.....+ 2017 . 2018 .2019 - 2017 . 2018 . 2016

=>A= 2017 . 2018 . 2019
 

ta có : \(S=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2017.2018}\)

\(\Leftrightarrow S=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)

\(\Leftrightarrow S=\dfrac{1}{2}-\dfrac{1}{2018}=\dfrac{504}{1009}\)

gọi biểu thức trên là A                                                                                                                                                                                          A=1/1 -1/2+1/3-1/4+...+1/2017-12018+1/2018-1/2019                                                                                                                                        A=1/1-1/2019                                                                                                                                                                                                       A=2018/2019

1/1.2+1/2.3+1/3.4+1/4.5+...+1/2017.2018+1/2018.2019

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\)

\(=1-\frac{1}{2019}\)

\(=\frac{2019}{2019}-\frac{1}{2019}\)

\(=\frac{2018}{2019}\)

cái ĐỒ ĐÁNG GHÉT ◥ὦɧ◤ŤŔầŃ VăŃ ĤùŃĞ™ kia t định trả lời sao m dám....

Các bạn giúp mk với. Mk đang cần gấp 😦

Trước tiên, chúng ta cần có lý thuyết về biến đổi phân số.

\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)

Ta có:

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

\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)

\(S=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+...-\dfrac{1}{2018}\)

\(S=1-\dfrac{1}{2018}\)

\(S=\dfrac{2017}{2018}\)

=1/1.2+1/2.3+1/3.4+...1/2017.2018

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2017-1/2018

=1-1/2018

=2018/2018-1/2018

=2017/2018

S = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\)+ .......+ \(\dfrac{1}{2017.2018}\)

S = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+.......+ \(\dfrac{1}{2017}\) - \(\dfrac{1}{2018}\)

S = \(\dfrac{1}{1}\) - \(\dfrac{1}{2018}\)

S = \(\dfrac{2017}{2018}\)