Đáp án:
\(\begin{array}{l}
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \dfrac{9}{2}; - \dfrac{1}{2}} \right) \cup \left( {\dfrac{1}{2}; + \infty } \right)\\
f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - \dfrac{9}{2}} \right) \cup \left( { - \dfrac{1}{2};\dfrac{1}{2}} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \pm \dfrac{1}{2}\\
f\left( x \right) = 0\\
\to \dfrac{{\left( { - 8{x^2} + x - 3} \right)\left( {2x + 9} \right)}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}} = 0\\
\to 2x + 9 = 0\\
\to x = - \dfrac{9}{2}\\
\left( {do: - 8{x^2} + x - 3 < 0\forall x} \right)
\end{array}\)
BXD:
x -∞ -9/2 -1/2 1/2 +∞
f(x) + 0 - // + // -
\(\begin{array}{l}
KL:f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \dfrac{9}{2}; - \dfrac{1}{2}} \right) \cup \left( {\dfrac{1}{2}; + \infty } \right)\\
f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - \dfrac{9}{2}} \right) \cup \left( { - \dfrac{1}{2};\dfrac{1}{2}} \right)
\end{array}\)
