\(\sqrt{x^2+4x+4}=\sqrt{x^2-6x+9}\)
\(\Leftrightarrow\sqrt{\left(x+2\right)^2}=\sqrt{\left(x-3\right)^2}\)
\(\Leftrightarrow\left|x+2\right|=\left|x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-3\\x+2=3-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2=-3\left(vl\right)\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2}\right\}\)
\(\sqrt{x^2+4x+4}=\sqrt{x^2-6x+9}\)
<=> \(\sqrt{\left(x+2\right)^2}=\sqrt{\left(x-3\right)^2}\)
<=> \(|x+2|=|x-3|\)
<=> \(\left[{}\begin{matrix}x+2=x-3\\x+2=-\left(x-3\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x-x=-3-2\\x+2=-x+3\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}0=-5\left(VLí\right)\\x+x=3-2\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-5\left(VLí\right)\\2x=1\end{matrix}\right.\)
<=> \(x=\dfrac{1}{2}\)
