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