Đáp án:
\(x \in \left( { - \infty ;\frac{{1 + \sqrt 7 }}{2}} \right]\)
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{\left| {{x^2} - 3x} \right| + 4 - {x^2} - x - 1}}{{{x^2} + x + 1}} \ge 0\\
\to \left| {{x^2} - 3x} \right| + 4 - {x^2} - x - 1 \ge 0\left( {do:{x^2} + x + 1 > 0\forall x \in R} \right)\\
\to \left| {{x^2} - 3x} \right| \ge {x^2} + x - 3\\
TH1:{x^2} - 3x \ge 0 \to x\left( {x - 3} \right) \ge 0\\
\to x \in \left( { - \infty ;0} \right] \cup \left[ {3; + \infty } \right)\left( 1 \right)\\
Bpt \to {x^2} - 3x \ge {x^2} + x - 3\\
\to 4x - 3 \le 0\\
\to x \le \frac{3}{4}\left( 2 \right)\\
\left( 1 \right);\left( 2 \right) \to x \in \left( { - \infty ;0} \right]\\
TH2:{x^2} - 3x < 0 \to x \in \left( {0;3} \right)\left( 3 \right)\\
Bpt \to - {x^2} + 3x \ge {x^2} + x - 3\\
\to 2{x^2} - 2x - 3 \le 0\\
\to x \in \left[ {\frac{{1 - \sqrt 7 }}{2};\frac{{1 + \sqrt 7 }}{2}} \right]\left( 4 \right)\\
\left( 3 \right);\left( 4 \right) \to x \in \left( {0;\frac{{1 + \sqrt 7 }}{2}} \right]\\
KL:x \in \left( { - \infty ;0} \right] \cup \left( {0;\frac{{1 + \sqrt 7 }}{2}} \right]\\
\to x \in \left( { - \infty ;\frac{{1 + \sqrt 7 }}{2}} \right]
\end{array}\)
