Đáp án:
c) \(x \in \left( { - \infty ; - 3} \right) \cup \left( {0;\dfrac{1}{2}} \right) \cup \left( {3; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 2;7} \right\}\\
\dfrac{{x - 5}}{{{x^2} - 5x - 14}} \ge 1\\
\to \dfrac{{x - 5 - {x^2} + 5x + 14}}{{\left( {x + 2} \right)\left( {x - 7} \right)}} \ge 0\\
\to \dfrac{{ - {x^2} + 6x + 9}}{{\left( {x + 2} \right)\left( {x - 7} \right)}} \ge 0\\
Xét: - {x^2} + 6x + 9 = 0\\
\to \left[ \begin{array}{l}
x = 3 + 3\sqrt 2 \\
x = 3 - 3\sqrt 2
\end{array} \right.
\end{array}\)
BXD:
x -∞ -2 3-3√2 7 3+3√2 +∞
f(x) - // + 0 - // + 0 -
\(KL:x \in \left[ { - 2;3 - 3\sqrt 2 } \right) \cup \left( {7;3 + 3\sqrt 2 } \right]\)
\(\begin{array}{l}
b)DK:x \ne \left\{ { - 3;0;3} \right\}\\
\dfrac{{x + 4}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{2}{{x + 3}} < \dfrac{{4x}}{{x\left( {3 - x} \right)}}\\
\to \dfrac{{{x^2} + 4x - 2\left( {{x^2} - 3x} \right) + 4x\left( {x + 3} \right)}}{{x\left( {x - 3} \right)\left( {x + 3} \right)}} < 0\\
\to \dfrac{{3{x^2} + 22x}}{{x\left( {x - 3} \right)\left( {x + 3} \right)}} < 0\\
\to \dfrac{{x\left( {3x + 22} \right)}}{{x\left( {x - 3} \right)\left( {x + 3} \right)}} < 0\\
\to \dfrac{{3x + 22}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} < 0
\end{array}\)
BXD:
x -∞ -22/3 -3 3 +∞
f(x) - 0 + // - // +
\(KL:x \in \left( { - \infty ; - \dfrac{{22}}{3}} \right) \cup \left( { - 3;3} \right)\)
\(\begin{array}{l}
c)DK:x \ne \dfrac{1}{2}\\
\dfrac{{x\left( {x - 3} \right)\left( {x + 3} \right)}}{{1 - 2x}} > 0
\end{array}\)
BXD:
x -∞ -3 0 1/2 3 +∞
f(x) - 0 + 0 - // + 0 -
\(KL:x \in \left( { - \infty ; - 3} \right) \cup \left( {0;\dfrac{1}{2}} \right) \cup \left( {3; + \infty } \right)\)
