Đáp án:
\[\left[ \begin{array}{l}
x = 0\\
x = 2\\
x = \dfrac{{5 \pm \sqrt {37} }}{2}
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge - 1\)
Ta có:
\(\begin{array}{l}
{x^2} + x + 3 = 3\sqrt {{x^3} + 1} \\
\Leftrightarrow {x^2} + x + 3 = 3.\sqrt {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)} \\
\Leftrightarrow \left( {{x^2} - x + 1} \right) + 2.\left( {x + 1} \right) = 3\sqrt {x + 1} .\sqrt {{x^2} - x + 1} \\
\Leftrightarrow \left( {{x^2} - x + 1} \right) - 3.\sqrt {x + 1} .\sqrt {{x^2} - x + 1} + 2.\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ {\left( {{x^2} - x + 1} \right) - \sqrt {x + 1} .\sqrt {{x^2} - x + 1} } \right] - \left[ {2.\sqrt {x + 1} .\sqrt {{x^2} - x + 1} - 2\left( {x + 1} \right)} \right] = 0\\
\Leftrightarrow \sqrt {{x^2} - x + 1} .\left( {\sqrt {{x^2} - x + 1} - \sqrt {x + 1} } \right) - 2\sqrt {x + 1} .\left( {\sqrt {{x^2} - x + 1} - \sqrt {x + 1} } \right) = 0\\
\Leftrightarrow \left( {\sqrt {{x^2} - x + 1} - \sqrt {x + 1} } \right)\left( {\sqrt {{x^2} - x + 1} - 2\sqrt {x + 1} } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {{x^2} - x + 1} = \sqrt {x + 1} \\
\sqrt {{x^2} - x + 1} = 2\sqrt {x + 1}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - x + 1 = x + 1\\
{x^2} - x + 1 = 4\left( {x + 1} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2x = 0\\
{x^2} - 5x - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = \dfrac{{5 \pm \sqrt {37} }}{2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\left( {t/m\,\,\,x \ge - 1} \right)
\end{array}\)
