Решите уравнение: корень из (5x-1)-корень из (3x-2)=1.

\[\sqrt{5x - 1} - \sqrt{3x - 2} = 1\]

\[u = \sqrt{5x - 1};\ \ \ v = \sqrt{3x - 2}\]

\[u - v = 1\]

\[3u^{2} - 5v^{2} =\]

\[= 3 \cdot (5x - 1) - 5 \cdot (3x - 2) =\]

\[\left\{ \begin{matrix} u - v = 1\ \ \ \ \ \ \ \ \ \\ 3u^{2} - 5v^{2} = 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} u = v + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3 \cdot (v + 1)^{2} - 5v^{2} = 7 \\ \end{matrix} \right.\ \]

\[3 \cdot \left( v^{2} + 2v + 1 \right) - 5v^{2} - 7 = 0\]

\[3v^{2} + 6v + 3 - 5v^{2} - 7 = 0\]

\[- 2v^{2} + 6v - 4 = 0\]

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

\[\left\{ \begin{matrix} v_{1} = 2 \\ u_{1} = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} v_{2} = 1 \\ u_{2} = 2 \\ \end{matrix} \right.\ \]

\[1)\ \left\{ \begin{matrix} \sqrt{3x - 2} = 2 \\ \sqrt{5x - 1} = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3x - 2 = 4 \\ 5x - 1 = 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3x = 6\ \ \\ 5x = 10 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 2 \\ x = 2 \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} \sqrt{3x - 2} = 1 \\ \sqrt{5x - 1} = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3x - 2 = 1 \\ 5x - 1 = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3x = 3 \\ 5x = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

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

\[Ответ:2;1.\ \]