$\left| x+\frac{3}{5} \right|-\left| x-\frac{7}{3} \right|=0$

$\left| x+\frac{3}{5} \right|=\left| x-\frac{7}{3} \right|$

$x+\frac{3}{5}=\pm \left( x-\frac{7}{3} \right)$

*TH 1 :

$x+\frac{3}{5}=-\left( x-\frac{7}{3} \right)$

$x+\frac{3}{5}=-x+\frac{7}{3}$

$2x=\frac{26}{15}$

$x=\frac{13}{15}$

*TH2 :

$x+\frac{3}{5}=x-\frac{7}{3}$

0 = $\frac{44}{15}$ ( vô lí $\Rightarrow $ loại )

Vậy $x=\frac{13}{15}$