Đáp án:
\(\dfrac{2}{3} > x\)
Giải thích các bước giải:
\(\begin{array}{l}
\left| {{x^2} - 3x + 4} \right| - 3x > {x^2}\\
\to \left| {{x^2} - 3x + 4} \right| > {x^2} + 3x\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 3x + 4 < - {x^2} - 3x\\
{x^2} - 3x + 4 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} - 3x + 4 > {x^2} + 3x\\
{x^2} - 3x + 4 \ge 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2{x^2} + 4 < 0\left( {vô lý} \right)\\
{x^2} - 3x + 4 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
4 > 6x\\
{x^2} - 3x + 4 \ge 0\left( {ld} \right)\forall x
\end{array} \right.
\end{array} \right.\\
\to \dfrac{2}{3} > x
\end{array}\)
