\(ĐK:x\in R\\ PT\Leftrightarrow\sqrt{\left(x+\dfrac{1}{2}\right)^2}=7-2x\\ \Leftrightarrow\left|x+\dfrac{1}{2}\right|=7-2x\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=7-2x,\forall x+\dfrac{1}{2}\ge0\\x+\dfrac{1}{2}=2x-7,\forall x+\dfrac{1}{2}< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{6},\forall x\ge-\dfrac{1}{2}\left(tm\right)\\x=\dfrac{15}{2},\forall x< -\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{13}{6}\)
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