\[\boxed{\text{465\ (465).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ 5\sqrt{x} = 3\]
\[\sqrt{x} = \frac{3}{5}\]
\[\left( \sqrt{x} \right)^{2} = \left( \frac{3}{5} \right)^{2}\]
\[\textbf{б)}\frac{1}{\sqrt{3x}} = 1\]
\[\sqrt{3x} = 1\ :1\]
\[\left( \sqrt{3x} \right)^{2} = (1)^{2}\]
\[3x = 1\]
\[\textbf{в)}\frac{1}{4\sqrt{x}} = 2\]
\[4\sqrt{x} = 1\ :2\]
\[4\sqrt{x} = \frac{1}{2}\]
\[\sqrt{x} = \frac{1}{2}\ :4 = \frac{1}{2} \cdot \frac{1}{4}\]
\[\sqrt{x} = \frac{1}{8}\]
\[\left( \sqrt{x} \right)^{2} = \left( \frac{1}{8} \right)^{2}\]
\[\textbf{г)}\ \sqrt{x - 5} = 4\]
\[\left( \sqrt{x - 5} \right)^{2} = 4^{2}\]
\[x - 5 = 16\]
\[\textbf{д)}\ 1 + \sqrt{2x} = 10\]
\[\sqrt{2x} = 9\]
\[\left( \sqrt{2x} \right)^{2} = 9^{2}\]
\[2x = 81\]
\[\textbf{е)}\ 3\sqrt{x} - 5 = 4\]
\[3\sqrt{x} = 4 + 5\]
\[3\sqrt{x} = 9\]
\[\sqrt{x} = 9\ :3\]
\[\sqrt{x} = 3\]
\[\left( \sqrt{x} \right)^{2} = 3^{2}\]