Đáp án:
3) \(x \in \left( {2;\dfrac{7}{2}} \right] \cup \left( {5; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
3)DK:x \ne \left\{ {2;5} \right\}\\
\dfrac{{2{x^2} - 16x + 27 - 2{x^2} + 14x - 20}}{{{x^2} - 7x + 10}} \le 0\\
\to \dfrac{{7 - 2x}}{{\left( {x - 2} \right)\left( {x - 5} \right)}} \le 0
\end{array}\)
BXD:
x -∞ 2 7/2 5 +∞
f(x) + // - 0 + // -
\(KL:x \in \left( {2;\dfrac{7}{2}} \right] \cup \left( {5; + \infty } \right)\)
\(\begin{array}{l}
4)DK:x \ne \left\{ { - 2; - \dfrac{4}{3};1;2} \right\}\\
\dfrac{{3{x^2} - 12 - 3{x^2} - x + 4}}{{\left( {3x + 4} \right)\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 2} \right)}} \ge 0\\
\to \dfrac{{ - x - 8}}{{\left( {3x + 4} \right)\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 2} \right)}} \ge 0
\end{array}\)
BXD:
x -∞ -8 -2 -4/3 1 2 +∞
f(x) + 0 - // + // - // + // -
\(KL:x \in \left[ { - 8;2} \right) \cup \left( { - \dfrac{4}{3};1} \right) \cup \left( {2; + \infty } \right)\)
