`|x+1|=|x-2|`
$\leftrightarrow \left[ \begin{array}{l}x+1=x-2\\x+1=-(x-2)\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}x\in∅\\x=\dfrac{1}{2}\end{array} \right.$
$\leftrightarrow x=\dfrac{1}{2}$
Vậy `x=1/2`
$\\$
`2|x-2|+|2-x|=15`
$\leftrightarrow 2|x-2|+|-(x-2)|=15$
$\leftrightarrow 2|x-2|+|x-2|=15$
$\leftrightarrow 3|x-2|=15$
$\leftrightarrow |x-2|=5$
$\leftrightarrow \left[ \begin{array}{l}x-2=5\\x-2=-5\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}x=7\\x=-3\end{array} \right.$
Vậy `x\in{-3;7}`
