(1.2 + 2.3 + 3.4 + ... + 2018.2019) - (1² + 2² + ... + 2018²)

= (1.2 + 2.3 + ... + 2018.2019) - (1.1 + 2.2 + ... + 2018.2018)

= (1.2 + 2.3 + ... + 2018.2019) - [1.(2 - 1) + 2.(3 - 1) + ... + 2018.(2019 - 1)]

= (1.2 + 2.3 + ... + 2018.2019) - (1.2 + 2.3 + ... + 2018.2019 - 1 - 2 - 3 - ... - 2018)

= (1.2 + 2.3 + ... + 2018.2019) - [1.2 + 2.3 + ... + 2018.2019 - (1 + 2 + ... + 2018)]

= (1.2 + 2.3 + ... + 2018.2019) - (1.2 + 2.3 + ... + 2018.2019) + (1 + 2 + 3 + ... + 2018)

= 1 + 2 + ... + 2018 (có : (2018 - 1) : 1 + 1 = 2018 (số))

= (2018 + 1).2018 : 2

= 2037171

cảm ơn nhé

cảm ơn 

Lời giải:
Ta có: \(x+(x+1)+(x+2)+...+(x+2017)=2017.2018\)

\(\Leftrightarrow \underbrace{(x+x+...+x)}_{2018}+(1+2+3+...+2017)=2017.2018\)

\(\Leftrightarrow 2018x+\frac{2017.2018}{2}=2017.2018\)

\(\Leftrightarrow 2018x=\frac{2017.2018}{2}\)

\(\Rightarrow x=\frac{2017}{2}\)

x = 2017 nha

Ta có : x(x-2017)-2018x+2017.2018=0

=>x(x-2017)-2018(x-2017)=0

=>(x-2017)(x-2018)=0

=>x=2017;2018.

`x/(1.2)+x/(2.3)+x/(3.4)+.....+x/(2017.2018)=1`

`-> x/1 - x/2 +x/2-x/3+x/3-x/4+........+x/2017-x/2018=1`

`-> x-x/2018=1`

`-> 2017/2018 .x=1`

`-> x=2018/2017`

x(1+2+3...2016)=2017.2018

 x.2017.2016:2=2017.2018

 1008x=2018

 x=1009/504

=> (x + x + x + .... + x) . (1 + 2 + 3 + ..... + 2016) = 2017.2018

=> 2016x . 1008 . 2017 = 2017 . 2018

=> 2016x . 1008 = 2018

=> /////////////

x+2x+3x+......+2016x=2017.2018

\(\Rightarrow\)x.(1+2+3+....+2016)=2017.2018

\(\Rightarrow\)x.2033136=2017.2018

\(\Rightarrow\)x=(2017.2018):2033136

x=\(\dfrac{1009}{504}\)

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