Điều kiện: \(x\ge2\)
\(x+2021\sqrt{x-2}=2\sqrt{x-1}\)
\(\Leftrightarrow x-2\sqrt{x-1}=-2021\sqrt{x-2}\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+1=-2021\sqrt{x-2}\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2-2021\sqrt{x-2}\)
Nhận xét:
\(VT=\left(\sqrt{x-1}-1\right)^2\ge0\forall x\ge2\)
\(VP=-2021\sqrt{x-2}\le0\forall x\ge2\)
\(VT=VP\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\-2021\sqrt{x-2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)
\(\Leftrightarrow x=2\left(TMĐK\right)\)
Vậy \(S=\left\{2\right\}\)
