Đặt \(x^2-2019=a\Rightarrow2019=x^2-a\) pt trở thành:
\(a^2+x=x^2-a\)
\(\Leftrightarrow a^2-x^2+x+a=0\)
\(\Leftrightarrow\left(a-x\right)\left(a+x\right)+a+x=0\)
\(\Leftrightarrow\left(a-x+1\right)\left(a+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2019-x+1=0\\x^2-2019+x=0\end{matrix}\right.\) \(\Rightarrow...\)