Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
4{x^2} - x + 1 < 0\\
4{x^2} - x + 1 = \left[ {{{\left( {2x} \right)}^2} - 2.2x.\frac{1}{4} + \frac{1}{{16}}} \right] + \frac{{15}}{{16}} = {\left( {2x - \frac{1}{4}} \right)^2} + \frac{{15}}{{16}} > 0,\,\,\,\forall x\\
\Rightarrow bpt\,\,\,4{x^2} - x + 1\,\,\,vn\\
b,\\
- 3{x^2} + x + 4 \ge 0\\
\Leftrightarrow 3{x^2} - x - 4 \le 0\\
\Leftrightarrow \left( {3x - 4} \right)\left( {x + 1} \right) \le 0\\
\Leftrightarrow - 1 \le x \le \frac{4}{3}\\
c,\\
DK:\,\,\,\,\,\left\{ \begin{array}{l}
x \ne \pm 2\\
x \ne 1\\
x \ne - \frac{4}{3}
\end{array} \right.\\
\frac{1}{{{x^2} - 4}} < \frac{3}{{3{x^2} + x - 4}}\\
\Leftrightarrow \frac{1}{{{x^2} - 4}} - \frac{3}{{3{x^2} + x - 4}} < 0\\
\Leftrightarrow \frac{{3{x^2} + x - 4 - 3.\left( {{x^2} - 4} \right)}}{{\left( {{x^2} - 4} \right)\left( {3{x^2} + x - 4} \right)}} < 0\\
\Leftrightarrow \frac{{x + 8}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 1} \right)\left( {3x + 4} \right)}} < 0\\
\Rightarrow S = \left( { - \infty ; - 8} \right) \cup \left( { - 2; - \frac{4}{3}} \right) \cup \left( {1;2} \right)\\
d,\\
{x^2} - x - 6 \le 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) \le 0\\
\Leftrightarrow - 2 \le x \le 3
\end{array}\)
