a/ \(0\le x\le2019^2\)
Đặt \(\sqrt{x}=t\ge0\Rightarrow t^2-2019+\sqrt{2019-t}=0\)
Đặt \(\sqrt{2019-t}=a\Rightarrow2019=a^2+t\) ta được:
\(t^2-\left(a^2+t\right)+a=0\)
\(\Leftrightarrow t^2-a^2-\left(t-a\right)=0\)
\(\Leftrightarrow\left(t-a\right)\left(t+a\right)-\left(t-a\right)=0\)
\(\Leftrightarrow\left(t-a\right)\left(t+a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=t\\a=1-t\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2019-t}=t\\\sqrt{2019-t}=1-t\left(t\le1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t^2+t-2019=0\\t^2-t-2018=0\end{matrix}\right.\) \(\Rightarrow t=...\Rightarrow x=t^2=...\)
