\(\frac{1}{3}\) + \(\frac{1}{6}\) + \(\frac{1}{10}\) + ... + \(\frac{1}{x\left(x+1\right):2}\)

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

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

= \(\frac{2016}{2018}=\frac{1008}{1009}\)

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2017}\)

\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2017}\)

\(\Leftrightarrow\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+..........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2018}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1008}{2018}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2018}\)

\(\Leftrightarrow x+1=2018\)

\(\Leftrightarrow x=2017\)

Vậy ..

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2018}\)

\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1008}{1009}\)

\(2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1008}{1009}\)

\(2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1008}{1009}\)

\(2.\left(\dfrac{1}{2}-\dfrac{1}{x-1}\right)\) = \(\dfrac{1008}{1009}\)

\(\dfrac{1}{2}-\dfrac{1}{x-1}=\dfrac{504}{1009}\)

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

\(x-1=2018\)

\(x=2019\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2016}{2018}\)

<=>  \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1008}{1009}\)

<=>  \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1008}{1009}\)

<=>   \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)

<=>  \(\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)

<=>  \(\frac{1}{x+1}=\frac{1}{2018}\)

=>  \(x+1=2018\)

<=>  \(x=2017\)

cảm ơn bạn

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.........+\dfrac{1}{x\left(x+1\right)}=\dfrac{2016}{2018}\)

\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2017}\)

\(\Leftrightarrow2\left(\dfrac{1}{6}+\dfrac{1}{12}+........+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2016}{2018}\)

\(\Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2016}{2018}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1008}{2018}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2018}\)

\(\Leftrightarrow x+1=2018\)

\(\Leftrightarrow x=2017\)

Vậy ...

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

Cảm ơn bạn rất nhiều mình đã hiểu rồi 

Chúc bạn học tốt nhé

Chỉ làm bài khó thôi nhé:::::::::::::::

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2016}{2018}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x.\left(x+1\right)}=\frac{2016}{2018}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2016}{2018}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{1013}{2018}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1013}{2018}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1013}{2018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2018}\Rightarrow x+1=2018\Rightarrow x=2017\)

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\cdot\left(x+1\right):2}=\dfrac{2016}{2018}\\ \dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\cdot\left(x+1\right)}=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}:2\\ \dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1008}{2018}\\ \dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{1008}{2018}\\ \dfrac{1}{x+1}=\dfrac{1}{2018}\\ \Leftrightarrow x+1=2018\\ x=2018-1\\ x=2017\)

cái đề j mk khó hỉu vậy viết đòang hoàng xem nào

Tìm số tự nhiên x biết

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\cdot\left(x+1\right):2}=\frac{2016}{2018}\)

ai làm được nào?

help me

\(a)\) Ta có : 

\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)

\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)

\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)

\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)

Lại có : 

\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)

\(\Rightarrow\)\(x=2019\)

Vậy \(x=2019\)

Chúc bạn học tốt ~ 

\(b)\) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(1-\frac{2}{x+1}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=1-\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2}{2019}\)

\(\Leftrightarrow\)\(x+1=2019\)

\(\Leftrightarrow\)\(x=2019-1\)

\(\Leftrightarrow\)\(x=2018\)

Vậy \(x=2018\)

Chúc bạn học tốt ~