ta có :
\(1.2+2.3+..+2017.2018=1^2+1+2^2+2+..+2017^2+2017\)
\(=\left(1^2+2^2+..+2017^2\right)+\left(1+2+..+2017\right)=\frac{2017.2018.\left(2017\times2+1\right)}{6}+\frac{2017.2018}{2}\)
\(=2017.2018.673\) nên vế trái bằng \(\frac{2017.2018.673}{2018.2019.x}=\frac{2017}{3x}\)
Xét vế trái \(=\frac{2}{2.3}+\frac{2}{3.4}+..+\frac{2}{2018.2019}=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{2018}-\frac{1}{2019}\right)=\frac{2.2017}{2.2019}=\frac{2017}{2019}\)
Vậy ta có : \(\frac{2017}{3x}=\frac{2017}{2019}\Leftrightarrow3x=2019\Leftrightarrow x=673\)
