\(3x-\sqrt{x^2+6x+9}=1\\ \Rightarrow3x-1=\sqrt{x^2+6x+9}\\ \Rightarrow3x-1=\left|x+3\right|\left(1\right)\)
TH1: \(x+3\ge0\Leftrightarrow x\ge-3\)
\(\left(1\right)\Leftrightarrow x+3=3x-1\Leftrightarrow2x=4\Leftrightarrow x=2\left(TM\right)\)
TH2: \(x+3< 0\Leftrightarrow x< -3\)
\(\left(1\right)\Leftrightarrow3x-1=-x-3\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\left(L\right)\)
Vậy \(x=2\)
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