Đáp án:
2) \(x \in \left[ { - 1;\dfrac{4}{3}} \right]\)
Giải thích các bước giải:
\(\begin{array}{l}
2)\\
\left( 1 \right):\left( {3{x^2} - x - 4} \right)\left( {{x^2} + 3} \right) \le 0\\
\to 3{x^2} - x - 4 \le 0\left( {do:{x^2} + 3 > 0\forall x} \right)\\
\to \left( {3x - 4} \right)\left( {x + 1} \right) \le 0
\end{array}\)
BXD:
x -∞ -1 4/3 +∞
f(x) + 0 - 0 +
\( \to x \in \left[ { - 1;\dfrac{4}{3}} \right]\)
\(\begin{array}{l}
\left( 2 \right)DK:x \ne 2\\
\dfrac{{5 - 2x}}{{{{\left( {x - 2} \right)}^2}}} \ge 0
\end{array}\)
BXD:
x -∞ 2 5/2 +∞
f(x) + // + 0 -
\(\begin{array}{l}
\to x \in \left( { - \infty ;2} \right) \cup \left( {2;\dfrac{5}{2}} \right]\\
KL:x \in \left[ { - 1;\dfrac{4}{3}} \right]
\end{array}\)
\(\begin{array}{l}
4)f\left( x \right) > 0\forall x\\
\to {x^2} - mx + m > 0\forall x\\
\to {m^2} - 4m < 0\\
\to m\left( {m - 4} \right) < 0\\
\to 0 < m < 4
\end{array}\)
