Решите уравнение: корень из (6x-2)-корень из (3x-2)=1.

\[\sqrt{6x - 2} - \sqrt{3x - 2} = 1\]

\[u = \sqrt{6x - 2};\ \ \ \ \ v = \sqrt{3x - 2}\]

\[u - v = 1\]

\[3u^{2} - 6v^{2} =\]

\[= 3 \cdot (6x - 2) - 6 \cdot (3x - 2) =\]

\[= 18x - 6 - 18 + 12 = 6\]

\[\left\{ \begin{matrix} u - v = 1\ \ \ \ \ \ \ \ \\ 3u^{2} - 6v^{2} = 6 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} u = v + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3 \cdot (v + 1)^{2} - 6v^{2} = 6 \\ \end{matrix} \right.\ \]

\[3 \cdot \left( v^{2} + 2v + 1 \right) - 6v^{2} = 6\]

\[3v^{2} + 6v + 3 - 6v^{2} - 6 = 0\]

\[- 3v^{2} + 6v - 3 = 0\]

\[v^{2} - 2v + 1 = 0\]

\[(v - 1)^{2} = 0\]

\[\left\{ \begin{matrix} u = v + 1\ \ \ \ \ \\ (v - 1)^{2} = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} v = 1 \\ u = 2 \\ \end{matrix} \right.\ \]

\[\ \left\{ \begin{matrix} \sqrt{3x - 2} = 1 \\ \sqrt{6x - 2} = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3x - 2 = 1 \\ 6x - 2 = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3x = 3 \\ 6x = 6 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 1 \\ x = 1 \\ \end{matrix} \right.\ \]

\[Ответ:\ 1.\]