Ta có :
\(\left|x-2,2\right|=\left|0,2+x\right|\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2,2=0,2+x\\x-2,2=-0,2+x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-x=0,2+2,2\\x+x=-0,2+2,2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0x=2,4\\2x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=-1\end{matrix}\right.\)
Vậy ................
Bài làm
Ta có : \(\left|x-2,2\right|=\left|0,2+x\right|\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2,2=0,2+x\\x-2,2=-0,2+x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-x=0,2+2,2\\x+x=-0,2+2,2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0x=2,4\\2x=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=-1\end{matrix}\right.\)
Vậy \(x=-1\)
