ĐKXĐ: \(x\ge0\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x\Leftrightarrow\left|x-3\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\left(x\ge3\right)\\x-3=-2x\left(0\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left|x-3\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\left(x\ge3\right)\\x-3=-2x\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
