Đáp án:
\(f\left( x \right) \ge 0\left( {ld} \right)\forall x \in R\)
Giải thích các bước giải:
\(\begin{array}{l}
f\left( x \right) = 2{x^4} - 2{x^3} - {x^2} + 1 \ge 0\\
\to 2{x^3}\left( {x - 1} \right) - {x^2} + x - x + 1 \ge 0\\
\to 2{x^3}\left( {x - 1} \right) - x\left( {x - 1} \right) - \left( {x - 1} \right) \ge 0\\
\to \left( {x - 1} \right)\left( {2{x^3} - x - 1} \right) \ge 0\\
\to \left( {x - 1} \right)\left( {x - 1} \right)\left( {2{x^2} + 2x + 1} \right) \ge 0\\
\to {\left( {x - 1} \right)^2}\left( {2{x^2} + 2x + 1} \right) \ge 0\\
Do:\left\{ \begin{array}{l}
{\left( {x - 1} \right)^2} \ge 0\forall x \in R\\
2{x^2} + 2x + 1 > 0\forall x \in R
\end{array} \right.\\
\to f\left( x \right) \ge 0\left( {ld} \right)\forall x \in R\\
\to dpcm
\end{array}\)
