\( \dfrac{1}{9}{\left( {x - 3} \right)^2} - \dfrac{1}{{25}}{\left( {x + 5} \right)^2} = 0\\ \Leftrightarrow 25{\left( {x - 3} \right)^2} - 9{\left( {x + 5} \right)^2} = 0\\ \Leftrightarrow \left[ {5\left( {x - 3} \right) - 3\left( {x + 5} \right)} \right]\left[ {5\left( {x - 3} \right) + 3\left( {x + 5} \right)} \right] = 0\\ \Leftrightarrow \left( {5x - 15 - 3x - 15} \right)\left( {5x - 15 + 3x + 15} \right) = 0\\ \Leftrightarrow \left( {2x - 30} \right).8x = 0\\ \Leftrightarrow 2\left( {x - 15} \right).8x = 0\\ \Leftrightarrow x\left( {x - 15} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x - 15 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 15 \end{array} \right. \)
\( {\left( {\dfrac{{3x}}{5} - \dfrac{1}{3}} \right)^2} = {\left( {\dfrac{x}{5} + \dfrac{2}{3}} \right)^2}\\ \Leftrightarrow \left| {\dfrac{{3x}}{5} - \dfrac{1}{3}} \right| = \left| {\dfrac{x}{5} + \dfrac{2}{3}} \right|\\ \Leftrightarrow \left[ \begin{array}{l} \dfrac{{3x}}{5} - \dfrac{1}{3} = \dfrac{x}{5} + \dfrac{2}{3}\\ \dfrac{{3x}}{5} - \dfrac{1}{3} = - \left( {\dfrac{x}{5} + \dfrac{2}{3}} \right) \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \dfrac{{2x}}{5} = 1\\ \dfrac{{4x}}{5} = - \dfrac{1}{3} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \dfrac{5}{2}\\ x = - \dfrac{5}{{12}} \end{array} \right. \)