Решите уравнение: корень из (5x-1)-корень из (4x-4)=1.

\[\sqrt{5x - 1} - \sqrt{4x - 4} = 1\]

\[u = \sqrt{5x - 1};\ \ \ \ v = \sqrt{4x - 4}\]

\[u - v = 1\]

\[4u^{2} - 5v^{2} =\]

\[= 4 \cdot (5x - 1) - 5 \cdot (4x - 4) =\]

\[= 20x - 4 - 20x + 20 = 16\ \ \]

\[\left\{ \begin{matrix} u - v = 1\ \ \ \ \ \ \ \ \ \ \ \\ 4u^{2} - 5v^{2} = 16 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} u = v + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4 \cdot (v + 1)^{2} - 5v^{2} = 16 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} u = v + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4v^{2} + 8v + 4 - 5v^{2} - 16 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} u = v + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - v^{2} + 8v - 12 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} u = v + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ v^{2} - 8v + 12 = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} v_{1} = 6 \\ u_{1} = 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

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

\[1)\ \left\{ \begin{matrix} \sqrt{4x - 4} = 6 \\ \sqrt{5x - 1} = 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 4x - 4 = 36 \\ 5x - 1 = 49 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 4x = 40 \\ 5x = 50 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 10 \\ x = 10 \\ \end{matrix} \right.\ \]

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

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

\[\left\{ \begin{matrix} 4x = 8\ \ \\ 5x = 10 \\ \end{matrix}\ \right.\ \text{\ \ \ \ \ \ }\]

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

\[Ответ:\ \ 10;2.\]