\[\boxed{\text{315\ (}\text{c}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\(\left( \sqrt{n^{2} + 39} \right)^{2} = x^{2}\)
\[n^{2} + 39 = x^{2}\]
\[x^{2} - n^{2} = 39\]
\[(x - n) \cdot (x + n) = 39\]
\[x - двузначное\ число;\ \ x \geq 0;\ \ \]
\[n \in N.\]
\[Ответ:при\ n = 19.\]
\[\boxed{\text{315\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[Преобразуем\ уравнения\ \]
\[согласно\ определению\ \]
\[арифметического\]
\[квадратного\ корня.\ \]
\[\textbf{а)}\ \sqrt{3x - 1} = 1\]
\[\left( \sqrt{3x - 1} \right)^{2} = 1^{2}\]
\[3x - 1 = 1\]
\[3x = 1 + 1\]
\[3x = 2\]
\[x = \frac{2}{3}.\]
\[\textbf{б)}\ \sqrt{6x + 4} = 2\]
\[\left( \sqrt{6x + 4} \right)^{2} = 2^{2}\]
\[6x + 4 = 4\]
\[6x = 0\]
\[x = 0.\]
\[\textbf{в)}\ \sqrt{12 - x} = 6\]
\[\left( \sqrt{12 - x} \right)^{2} = 6^{2}\]
\[12 - x = 36\]
\[x = 12 - 36\]
\[x = - 24.\]
\[\textbf{г)}\ \sqrt{8x - 1} = 1\]
\[\left( \sqrt{8x - 1} \right)^{2} = 1^{2}\]
\[8x - 1 = 1\]
\[8x = 2\]
\[x = \frac{2}{8} = \frac{1}{4}\]
\[x = 0,25.\]