Đáp án:
q. \(\left[ \begin{array}{l}
x \ge \dfrac{{5 + \sqrt {13} }}{2}\\
x \le \dfrac{{5 - \sqrt {13} }}{2};x \ne - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
h.\dfrac{{\left( {4x - 1} \right)\left( {x + 1} \right)}}{{{x^2} + 5x + 7}} > 0\\
\to \left( {4x - 1} \right)\left( {x + 1} \right) > 0\left( {do:{x^2} + 5x + 7 > 0\forall x \in R} \right)\\
\to \left[ \begin{array}{l}
x > \dfrac{1}{4}\\
x < - 1
\end{array} \right.\\
m.\left( {x + 3} \right)\left( {x + 2} \right)\left( {x + 1} \right) > 0
\end{array}\)
BXD
x -∞ -3 -2 -1 +∞
y - 0 + 0 - 0 +
\(KL:x \in \left( { - 3; - 2} \right) \cup \left( { - 1; + \infty } \right)\)
\(\begin{array}{l}
n.{x^2} - 3x + 2 > 0\\
\to \left( {x - 2} \right)\left( {x - 1} \right) > 0\\
\to \left[ \begin{array}{l}
x > 2\\
x < 1
\end{array} \right.\\
p.DK:x \ne \pm 1\\
\dfrac{{{x^2} - x - 6 - {x^2} + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} < 0\\
\to \dfrac{{ - x - 5}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} < 0
\end{array}\)
BXD:
x -∞ -5 -1 1 +∞
y + 0 - // + // -
\(KL:x \in \left( { - 5; - 1} \right) \cup \left( {1; + \infty } \right)\)
\(\begin{array}{l}
q.DK:x \ne \left\{ { - 1;2} \right\}\\
\dfrac{1}{{{{\left( {x - 2} \right)}^2}}} \le \dfrac{1}{{x + 1}}\\
\to {\left( {x - 2} \right)^2} \ge x + 1\\
\to {x^2} - 4x + 4 \ge x + 1\\
\to {x^2} - 5x + 3 \ge 0\\
\to \left[ \begin{array}{l}
x \ge \dfrac{{5 + \sqrt {13} }}{2}\\
x \le \dfrac{{5 - \sqrt {13} }}{2}
\end{array} \right.
\end{array}\)\
\(KL:\left[ \begin{array}{l}
x \ge \dfrac{{5 + \sqrt {13} }}{2}\\
x \le \dfrac{{5 - \sqrt {13} }}{2};x \ne - 1
\end{array} \right.\)
