\(7-5\left|2-x\right|=0\)
\(\Leftrightarrow5\left|2-x\right|=7\)
\(\Leftrightarrow\left|2-x\right|=\dfrac{7}{5}\)
\(TH_1:x\le2\)
\(2-x=\dfrac{7}{5}\)
\(\Leftrightarrow x=\dfrac{3}{5}\left(tm\right)\)
\(TH_2:x\ge2\)
\(2-x=-\dfrac{7}{5}\)
\(\Leftrightarrow x=\dfrac{17}{5}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{3}{5};\dfrac{17}{5}\right\}\)