Số nghiệm của phương trình \(\sqrt {5x - 1}  = \sqrt {3x - 2}  + \sqrt {x - 1} \) là

\(\sqrt {5x - 1}  = \sqrt {3x - 2}  + \sqrt {x - 1} \)

TXĐ: \(D = \left[ {1; + \infty } \right)\)

\(\begin{array}{l}\,\,\,\,\,\,\,\,\sqrt {5x - 1}  = \sqrt {3x - 2}  + \sqrt {x - 1} \\ \Leftrightarrow {\left( {\sqrt {5x - 1} } \right)^2} = {\left( {\sqrt {3x - 2}  + \sqrt {x - 1} } \right)^2}\\ \Leftrightarrow 5x - 1 = 3x - 2 + x - 1 + 2\sqrt {\left( {3x - 2} \right)\left( {x - 1} \right)} \\ \Leftrightarrow x + 2 = 2\sqrt {3{x^2} - 5x + 2} \\ \Leftrightarrow \left\{ \begin{array}{l}x + 2 \ge 0\\{\left( {x + 2} \right)^2} = 4\left( {3{x^2} - 5x + 2} \right)\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}x \ge  - 2\\{x^2} + 4x + 4 = 12{x^2} - 20x + 8\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge  - 2\\11{x^2} - 24x + 4 = 0\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}x \ge  - 2\\\left[ \begin{array}{l}x = 2\\x = \dfrac{2}{{11}}\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 2\,\,\left( {tm} \right)\\x = \dfrac{2}{{11}}\,\,\left( {ktm} \right)\end{array} \right.\end{array}\)

Vậy phương trình có 1 nghiệm \(x = 2\)