Đáp án:
x=1
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{{{x^3} + x}}{{{{\left( {{x^2} - x + 1} \right)}^2}}} = 2\\
\to {x^3} + x = 2{\left( {{x^2} - x + 1} \right)^2}\\
\to {x^3} + x = 2\left( {{x^4} + {x^2} + 1 - 2{x^3} + 2{x^2} - 2x} \right)\\
\to {x^3} + x = 2{x^4} + 2{x^2} + 2 - 4{x^3} + 4{x^2} - 4x\\
\to 2{x^4} - 5{x^3} + 6{x^2} - 5x + 2 = 0\\
\to 2{x^4} - 2{x^3} - 3{x^3} + 3{x^2} + 3{x^2} - 3x - 2x + 2 = 0\\
\to 2{x^3}\left( {x - 1} \right) - 3{x^2}\left( {x - 1} \right) + 3x\left( {x - 1} \right) - 2\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {2{x^3} - 3{x^2} + 3x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
2{x^3} - 2{x^2} - {x^2} + x + 2x - 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
2{x^2}\left( {x - 1} \right) - x\left( {x - 1} \right) + 2\left( {x - 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
\left( {x - 1} \right)\left( {2{x^2} - x + 2} \right) = 0
\end{array} \right.\\
\to x = 1\left( {do:2{x^2} - x + 2 > 0\forall x} \right)
\end{array}\)
