Giải thích các bước giải:
\(\begin{array}{l}
a.f\left( x \right) \ge 0 \Leftrightarrow - 3{x^2} + x + 4 \ge 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {3x - 4} \right) \ge 0\\
Xét:\left[ \begin{array}{l}
x + 1 = 0\\
3x - 4 = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = - 1\\
x = \frac{4}{3}
\end{array} \right.
\end{array}\)
BXD:
x -∞ -1 4/3 +∞
f(x) - 0 + 0 -
\(KL:x \in \left[ { - 1;\frac{4}{3}} \right]\)
\(\begin{array}{l}
b.DK:\left\{ \begin{array}{l}
{x^2} - 4 \ne 0\\
3{x^2} + x - 4 \ne 0
\end{array} \right. \to \left\{ \begin{array}{l}
x \ne \pm 2\\
x \ne 1\\
x \ne - \frac{4}{3}
\end{array} \right.\\
f\left( x \right) = \frac{1}{{{x^2} - 4}} - \frac{3}{{3{x^2} + x - 4}} < 0\\
\to \frac{{3{x^2} + x - 4 - 3{x^2} + 12}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 1} \right)\left( {3x + 4} \right)}} < 0\\
\to \frac{{x + 8}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 1} \right)\left( {3x + 4} \right)}} < 0
\end{array}\)
BXD:
x -∞ -2 -4/3 1 2 8 +∞
f(x) - // + // - // + // - 0 +
\(KL:x \in \left( { - \infty ; - 2} \right) \cup \left( { - \frac{4}{3};1} \right) \cup \left( {2;8} \right)\)
