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

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
 

a) = 1-1/2+1/2-1/3+1/3-1/4

    = 1-1/4=3/4

b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018

   =1-1/2018=2017/2018

c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015

   = 1/2-1/2015=2015/4030-2/4030=2013/4030

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)

b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

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

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

\(=\frac{2017}{2018}\)

c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)

\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)

\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)

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

\(=\frac{3}{2}.\frac{2013}{4030}\)

\(=\frac{6039}{8060}\)

]\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

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

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

             \(=\frac{3}{1}-\frac{3}{2018}=\frac{6051}{2018}\)

Vậy \(\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+\frac{3}{3\cdot4}+...+\frac{3}{2017\cdot2018}=\frac{6051}{2018}\)

3/1.2 + 3/2.3 + 3/3.4 + ... + 3/2017.2018

= \(3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)

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

= 3 . ( 1 - 1/2018 )

= 3 . 2017/2018

= 6051/2018

                            giải

3/1.2+3/2.3+3/3.4+.....+3/2017.2018

=3/1+3/2018

=6055/2018 

k mình nhé