Đáp án:
\[\left[ \begin{array}{l}
x = 2\\
x = 1
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge 1\)
Ta có:
\(\begin{array}{l}
\left( {\sqrt {5x - 1} + \sqrt {x - 1} } \right)\left( {3x - 1 - \sqrt {5{x^2} - 6x + 1} } \right) = 4x\\
\Leftrightarrow \left( {\sqrt {5x - 1} + \sqrt {x - 1} } \right)\left( {6x - 2 - 2\sqrt {5{x^2} - 6x + 1} } \right) = 2.\left[ {\left( {5x - 1} \right) - \left( {x - 1} \right)} \right]\\
\Leftrightarrow \left( {\sqrt {5x - 1} + \sqrt {x - 1} } \right)\left[ {\left( {5x - 1} \right) - 2\sqrt {\left( {5x - 1} \right)\left( {x - 1} \right)} + \left( {x - 1} \right)} \right] = 2\left( {\sqrt {5x - 1} - \sqrt {x - 1} } \right)\left( {\sqrt {5x - 1} + \sqrt {x - 1} } \right)\\
\Leftrightarrow \left( {\sqrt {5x - 1} + \sqrt {x - 1} } \right){\left( {\sqrt {5x - 1} - \sqrt {x - 1} } \right)^2} = 2\left( {\sqrt {5x - 1} + \sqrt {x - 1} } \right)\left( {\sqrt {5x - 1} - \sqrt {x - 1} } \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {5x - 1} + \sqrt {x - 1} = 0\\
\sqrt {5x - 1} - \sqrt {x - 1} = 0\\
\sqrt {5x - 1} - \sqrt {x - 1} = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
VN\\
5x - 1 = x - 1\\
6x - 2 - 2\sqrt {\left( {5x - 1} \right)\left( {x - 1} \right)} = 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\left( L \right)\\
\sqrt {\left( {5x - 1} \right)\left( {x - 1} \right)} = 3x - 3
\end{array} \right. \Leftrightarrow 5{x^2} - 6x + 1 = 9{x^2} - 18x + 9\\
\Leftrightarrow 4{x^2} - 12x + 8 = 0 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 1
\end{array} \right.\left( {t/m} \right)
\end{array}\)
