ĐKXĐ : \(\hept{\begin{cases}x^2+6x+9\ge0\\3x-6\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+3\right)^2\ge0\\x\ge2\end{cases}\Rightarrow}x\ge2}\)
\(\sqrt{x^2+6x+9}=3x-6\)
\(\Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-6\)
\(\Leftrightarrow\left|x+3\right|=3x-6\)
Ta có : \(\left|x+3\right|=\hept{\begin{cases}x+3\Leftrightarrow x\ge-3\\-x-3\Leftrightarrow x< -3\left(KTMĐKĐ\right)\end{cases}}\)
Xét \(x\ge2\) thì \(x+3=3x-6\Leftrightarrow x-3x=-6-3\Leftrightarrow-2x=-9\Rightarrow x=\frac{9}{2}\)(TM)
Vậy \(x=\frac{9}{2}\)
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