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