Tìm x biết 1018x^2 + x - 2019 = 0

* Hướng dẫn giải

$\begin{array}{l}2018{\mathrm{x}}^2+\mathrm{x}-2019=0\\ 2018{\mathrm{x}}^2-2018\mathrm{x}+2019\mathrm{x}-2019=0\\ 2018\mathrm{x}\left(\mathrm{x}-1\right)+2019\left(\mathrm{x}-1\right)=0\\ \left(\mathrm{x}-1\right)\left(2018\mathrm{x}+2019\right)=0\\ \iff \left[\begin{array}{l}\mathrm{x}=1\\ \mathrm{x}=-\frac{2019}{2019}\end{array}\right.\end{array}$