Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = {x^4} - {x^2} + 6x - 9\\
= {x^4} - \left( {{x^2} - 6x + 9} \right)\\
= {x^4} - {\left( {x - 3} \right)^2}\\
= \left( {{x^2} - x + 3} \right)\left( {{x^2} + x - 3} \right)\\
{x^2} - x + 3 = \left( {{x^2} - x + \frac{1}{4}} \right) + \frac{{11}}{4} = {\left( {x - \frac{1}{2}} \right)^2} + \frac{{11}}{4} > 0,\,\,\,\,\forall x\\
f\left( x \right) \ge 0 \Leftrightarrow {x^2} + x - 3 \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \ge \frac{{ - 1 + \sqrt {13} }}{2}\\
x \le \frac{{ - 1 - \sqrt {13} }}{2}
\end{array} \right.\\
f\left( x \right) < 0 \Leftrightarrow {x^2} + x - 3 < 0 \Leftrightarrow \frac{{ - 1 - \sqrt {13} }}{2} < x < \frac{{ - 1 + \sqrt {13} }}{2}
\end{array}\)
