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