Các bạn giúp mk với. Mk đang cần gấp 😦

Ta có : 

\(A=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{1.2+2.3+3.4+...+2017.2018}\)

\(A=\frac{\frac{2}{2}+\frac{2\left(2+1\right)}{2}+\frac{3\left(3+1\right)}{2}+...+\frac{2017\left(2017+1\right)}{2}}{1.2+2.3+3.4+...+2017.2018}\)

\(A=\frac{\frac{2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{2017.2018}{2}}{1.2+2.3+3.4+...+2017.2018}\)

\(A=\frac{\frac{1.2+2.3+3.4+...+2017.2018}{2}}{1.2+2.3+3.4+...+2017.2018}\)

\(A=\frac{1.2+2.3+3.4+...+2017.2018}{2}.\frac{1}{1.2+2.3+3.4+...+2017.2018}\)

\(A=\frac{1}{2}\)

Vậy \(A=\frac{1}{2}\)

Chúc bạn học tốt ~ 

Cảm ơn PMQ nhiều nha cậu cứu mình rồi

ai tích mình mình tích lại cho

tinh tong A= 1.2^2+2.3^2+3.4^2+........+2018.2019^2

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)

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

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

\(=2.\frac{2017}{2018}\)

\(=\frac{2017}{1009}\)

quy tử số thành 1 

A = 2.(1/1.2+1/3.2+1/3.4+... + 1/2017.2018) 

A= 2. (1- 1/2018) 

Tính nốt nha

A=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{2017.2018}\)

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

A=\(2-\frac{2}{2018}\)

A=\(\frac{1017}{509}\)

Tk cho mk mk tk lại

sorry mọi ng nha,toán 6 chứ ko phải toán 5

(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