$15)$
$|x+2|+|x+\dfrac{3}{5}|+|x+\dfrac{1}{2}|=4x$
Vì: $\begin{cases} |x+2|≥0∀x\\|x+\dfrac{3}{5}|≥0∀x\\|x+\dfrac{1}{2}|≥0∀x \end{cases}$
$⇒|x+2|+|x+\dfrac{3}{5}|+|x+\dfrac{1}{2}|≥0∀x$
$⇒4x≥0⇔x≥0$
Dấu "=" xảy ra $⇔x+2+x+\dfrac{3}{5}+x+\dfrac{1}{2}=4x$
$⇔x+x+x-4x=-2-\dfrac{3}{5}-\dfrac{1}{2}$
$⇔-x=-\dfrac{31}{10}$
$⇔x=\dfrac{31}{10}$
Vậy: $x=\dfrac{31}{10}$
