$g$) `|x + 4/{15}| - |-3,75| = - |-2,15|`
`⇔ |x+ 4/{15}| = {8}/{5}`
`⇒` \(\left[ \begin{array}{l}x=\dfrac{4}{3}\\x=-\dfrac{-28}{15}\end{array} \right.\)
Vậy $x$ $∈$ `{{4}/{3};{-28}/{15}}`
$h$) `|2x+1|=|x|`
`⇒` \(\left[ \begin{array}{l}2x+1=x\\2x+1=-x\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=-1\\x=\dfrac{-1}{3}\end{array} \right.\)
Vậy $x$ $∈$ `{-1;-1/3}`
