Ta có: \(\frac{1}{2}+\frac{1}{2}\left|x+2\right|=10\)(1)
Trường hợp 1: \(x\ge-2\)
(1)\(\Leftrightarrow\)\(\frac{1}{2}+\frac{1}{2}\left(x+2\right)=10\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}x+1-10=0\)
\(\Leftrightarrow\frac{1}{2}x-\frac{17}{12}=0\)
\(\Leftrightarrow\frac{1}{2}x=\frac{17}{2}\)
hay \(x=\frac{17}{12}:\frac{1}{2}=\frac{17}{2}\cdot2=17\)(tm)
*Trường hợp 2: x<-2
(1)\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}\left(-x-2\right)=10\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}x-1-10=0\)
\(\Leftrightarrow\frac{-1}{2}x-\frac{21}{2}=0\)
\(\Leftrightarrow\frac{-1}{2}x=\frac{21}{2}\)
hay \(x=\frac{21}{2}:\frac{-1}{2}=\frac{21}{2}\cdot\left(-2\right)=-21\)(tm)
Vậy: S={17;-21}
