Đáp án:
\[\left[ \begin{array}{l}
x = 1\\
x = \frac{{3 \pm \sqrt {21} }}{2}\\
x = 4 \pm 3\sqrt 2
\end{array} \right.\]
Giải thích các bước giải:
33,
ĐKXĐ: \(x \ge - \sqrt[3]{3}\)
Ta có:
\(\begin{array}{l}
{x^3} + 6{x^2} - 2x + 3 = \left( {5x - 1} \right)\sqrt {{x^3} + 3} \\
\Leftrightarrow \left( {{x^3} + 3} \right) + 2x\left( {3x - 1} \right) = \left[ {2x + \left( {3x - 1} \right)} \right]\sqrt {{x^3} + 3} \\
\Leftrightarrow \left( {{x^3} + 3} \right) + 2x\left( {3x - 1} \right) - 2x\sqrt {{x^3} + 3} - \left( {3x - 1} \right)\sqrt {{x^3} + 3} = 0\\
\Leftrightarrow \sqrt {{x^3} + 3} \left( {\sqrt {{x^3} + 3} - 2x} \right) + \left( {3x - 1} \right)\left[ {2x - \sqrt {{x^3} + 3} } \right] = 0\\
\Leftrightarrow \left( {\sqrt {{x^3} + 3} - 2x} \right)\left( {\sqrt {{x^3} + 3} - \left( {3x - 1} \right)} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {{x^3} + 3} = 2x\\
\sqrt {{x^3} + 3} = 3x - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{x^3} + 3 = 4{x^2}\\
{x^3} + 3 = 9{x^2} - 6x + 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^3} - 4{x^2} + 3 = 0\\
{x^3} - 9{x^2} + 6x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{{3 \pm \sqrt {21} }}{2}\\
x = 4 \pm 3\sqrt 2
\end{array} \right.\left( {t/m} \right)
\end{array}\)
