Đáp án:
\[\left[ {\begin{array}{*{20}{c}}
{x = 1}\\
{x = 2}
\end{array}} \right.\]
Giải thích các bước giải:
\(\begin{array}{l}
{x^3} + 3{x^2} + 4x + 2 = 5x\sqrt {5x - 1} \\
dk:x \ge \frac{1}{5}\\
pt \Leftrightarrow {x^3} + 3{x^2} - 6x + 2 = 5x\sqrt {5x - 1} - 10x\\
\Leftrightarrow {x^3} - {x^2} + 4{x^2} - 4x - 2x + 2 = 5x(\sqrt {5x - 1} - 2)\\
\Leftrightarrow {x^2}(x - 1) + 4x(x - 1) - 2(x - 1) = 5x.\frac{{5x - 5}}{{\sqrt {5x - 1} + 2}}\\
\Leftrightarrow (x - 1)({x^2} + 4x - 2) - \frac{{25x(x - 1)}}{{\sqrt {5x - 1} + 2}} = 0\\
\Leftrightarrow (x - 1)({x^2} + 4x - 2 - \frac{{25x}}{{\sqrt {5x - 1} + 2}}) = 0\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{x = 1}\\
{{x^2} + 4x - 2 - \frac{{25x}}{{\sqrt {5x - 1} + 2}} = 0(2)}
\end{array}} \right.\\
(2) \Leftrightarrow {x^2} - 2x + 6x - 12 + 10 - \frac{{25x(\sqrt {5x - 1} - 2)}}{{5x - 5}} = 0\\
\Leftrightarrow x(x - 2) + 6(x - 2) + 10 - \frac{{5x\sqrt {5x - 1} - 10x}}{{x - 1}} = 0\\
\Leftrightarrow (x - 2)(x + 6) + \frac{{15x - 5x\sqrt {5x - 1} + 5x - 10}}{{x - 1}} = 0\\
\Leftrightarrow (x - 2)(x + 6) + \frac{{5x(3 - \sqrt {5x - 1} ) + 5(x - 2)}}{{x - 1}} = 0\\
\Leftrightarrow (x - 2)(x + 6) + \frac{{5x.\frac{{10 - 5x}}{{3 + \sqrt {5x - 1} }} + 5(x - 2)}}{{x - 1}} = 0\\
\Leftrightarrow (x - 2)(x + 6) + \frac{{(x - 2)(\frac{{ - 25x}}{{3 + \sqrt {5x - 1} }} + 5)}}{{x - 1}} = 0\\
\Leftrightarrow (x - 2)(x + 6 + \frac{{ - 25x + 15 + 5\sqrt {5x - 1} }}{{(x - 1)(3 + \sqrt {5x - 1} )}}) = 0\\
\Leftrightarrow x = 2
\end{array}\)
