\[\boxed{\text{314\ (314).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[Преобразуем\ уравнения\ \]
\[согласно\ определению\ \]
\[арифметического\]
\[квадратного\ корня.\ \]
\[\textbf{а)}\ \sqrt{3 + 5x} = 7\ \]
\[\left( \sqrt{3 + 5x} \right)^{2} = 7^{2}\]
\[3 + 5x = 49\]
\[5x = 46\]
\[x = \frac{46}{5}\]
\[x = 9,2.\]
\[\textbf{б)}\ \sqrt{10x - 14} = 11\ \]
\[\left( \sqrt{10x - 14} \right)^{2} = 11^{2}\]
\[10x - 14 = 121\]
\[10x = 135\]
\[x = \frac{135}{10}\]
\[x = 13,5.\]
\[\textbf{в)}\ \sqrt{\frac{1}{3}x - \frac{1}{2}} = 0\]
\[\left( \sqrt{\frac{1}{3}x - \frac{1}{2}} \right)^{2} = 0^{2}\]
\[\frac{1}{3}x - \frac{1}{2} = 0\]
\[\frac{1}{3}x = \frac{1}{2}\]
\[x = \frac{1}{2}\ :\frac{1}{3} = \frac{1}{2} \cdot 3\]
\[x = \frac{3}{2} = 1,5.\]