`c)`
`|x-1,5|+|2,5-x|=0`
$⇒\begin{cases} x-1,5=0\\2,5-x=0 \end{cases}$
$⇒\begin{cases} x=1,5\\x=2,5 \end{cases}$
Vậy `x∈{1,5;2,5}`
`d)`
`|x-5|=|2x+1|`
`=>`\(\left[ \begin{array}{l}x-5=2x+1\\x-5=-2x-1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=-6\\x=\dfrac{4}{3}\end{array} \right.\)
Vậy `x∈{-6;4/3}`