Đáp án:
\(x \in \left( { - \infty ; - 9} \right) \cup \left[ { - 6; - 3} \right] \cup \left( {0; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ { - 9;0} \right\}\\
\dfrac{1}{{x + 9}} - \dfrac{1}{x} - \dfrac{1}{2} \le 0\\
\to \dfrac{{2x - 2\left( {x + 9} \right) - x\left( {x + 9} \right)}}{{2x\left( {x + 9} \right)}} \le 0\\
\to \dfrac{{2x - 2x - 18 - {x^2} - 9x}}{{2x\left( {x + 9} \right)}} \le 0\\
\to \dfrac{{ - {x^2} - 9x - 18}}{{2x\left( {x + 9} \right)}} \le 0\\
\to \dfrac{{ - \left( {x + 3} \right)\left( {x + 6} \right)}}{{2x\left( {x + 9} \right)}} \le 0
\end{array}\)
BXD:
x -∞ -9 -6 -3 0 +∞
f(x) - // + 0 - 0 + // -
\(KL:x \in \left( { - \infty ; - 9} \right) \cup \left[ { - 6; - 3} \right] \cup \left( {0; + \infty } \right)\)
