ĐKXĐ : \(x\ge-\dfrac{7}{3}\)
\(x^2+7x+12=2\sqrt{3x+7}\)
\(\Leftrightarrow x^2+7x+12-2\sqrt{3x+7}=0\)
\(\Leftrightarrow\left(x^2+4x+4\right)+\left(3x+7-2\sqrt{3x+7}+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2+\left(\sqrt{3x+7}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)^2=0\\\left(\sqrt{3x+7}-1\right)^2=0\end{matrix}\right.\Leftrightarrow x=-2\left(TMĐK\right)\)
Vậy \(S=\left\{-2\right\}\)
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