\(\Leftrightarrow\sqrt{\left(x-3\right)^2+1}=1-2\left(x-3\right)^2\)
Do \(\left(x-3\right)^2\ge0\Rightarrow\left\{{}\begin{matrix}VT=\sqrt{\left(x-3\right)^2+1}\ge1\\VP=1-\left(x-3\right)^2\le1\end{matrix}\right.\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi \(\left(x-3\right)^2=0\Leftrightarrow x=3\)