Đáp án:
a) \(x \in \left( { - \infty ; - 1} \right] \cup \left[ {\dfrac{3}{5};1} \right] \cup \left[ {2; + \infty } \right)\)
Giải thích các bước giải:
\(a)({x^2} - 1)(4 - 2x)(5x - 3) \le 0\)
BXD:
x -∞ -1 3/5 1 2 +∞
f(x) - 0 + 0 - 0 + 0 -
\(KL:x \in \left( { - \infty ; - 1} \right] \cup \left[ {\dfrac{3}{5};1} \right] \cup \left[ {2; + \infty } \right)\)
\(\begin{array}{l}
b)DK:x \ne 4\\
\dfrac{{\left( {{x^2} - 2x + 1} \right)\left( {3 - x} \right)}}{{4x - 16}} \le 0\\
\to \dfrac{{{{\left( {x - 1} \right)}^2}\left( {3 - x} \right)}}{{4\left( {x - 4} \right)}} \le 0
\end{array}\)
BXD:
x -∞ 1(kép) 3 4 +∞
f(x) - 0 - 0 + // -
\(KL:x \in \left( { - \infty ;3} \right] \cup \left( {4; + \infty } \right)\)
\(\begin{array}{l}
c)DK:x \ne \left\{ { - 4; - 3;0} \right\}\\
\dfrac{1}{x} + \dfrac{2}{{x + 4}} < \dfrac{3}{{x + 3}}\\
\to \dfrac{{{x^2} + 7x + 12 + 2{x^2} + 6x - 3{x^2} - 12x}}{{x\left( {x + 3} \right)\left( {x + 4} \right)}} < 0\\
\to \dfrac{{x + 12}}{{x\left( {x + 3} \right)\left( {x + 4} \right)}} < 0
\end{array}\)
BXD:
x -∞ -12 -4 -3 0 +∞
f(x) + 0 - // + // - // +
\(KL:x \in \left( { - 12; - 4} \right) \cup \left( { - 3;0} \right)\)
