Đáp án:
20) \(x \in \left( { - \infty ; - \dfrac{1}{2}} \right) \cup \left( {1;3} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
20)DK:x \ne \left\{ {1;3} \right\}\\
\dfrac{{x + 2}}{{x - 1}} > \dfrac{{x + 4}}{{x - 3}}\\
\to \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right) - \left( {x - 1} \right)\left( {x + 4} \right)}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} > 0\\
\to \dfrac{{{x^2} - x - 6 - {x^2} - 3x + 4}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} > 0\\
\to \dfrac{{ - 4x - 2}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} > 0
\end{array}\)
BXD:
x -∞ -1/2 1 3 +∞
f(x) + 0 - // + // -
\(KL:x \in \left( { - \infty ; - \dfrac{1}{2}} \right) \cup \left( {1;3} \right)\)
\(\begin{array}{l}
21)DK:x \ne \left\{ { - 4; - 3;0} \right\}\\
\dfrac{1}{x} + \dfrac{2}{{x + 4}} < \dfrac{3}{{x + 3}}\\
\to \dfrac{{\left( {x + 4} \right)\left( {x + 3} \right) + 2x\left( {x + 3} \right) - 3x\left( {x + 4} \right)}}{{x\left( {x + 4} \right)\left( {x + 3} \right)}} < 0\\
\to \dfrac{{{x^2} + 7x + 12 + 2{x^2} + 6x - 3{x^2} - 12x}}{{x\left( {x + 4} \right)\left( {x + 3} \right)}} < 0\\
\to \dfrac{{x + 12}}{{x\left( {x + 4} \right)\left( {x + 3} \right)}} < 0
\end{array}\)
BXD:
x -∞ -12 -4 -3 0 +∞
f(x) + 0 - // + // - // +
\(KL:x \in \left( { - 12; - 4} \right) \cup \left( { - 3;0} \right)\)
\(\begin{array}{l}
22)DK:x \ne \left\{ { - 3; - 2; - 1} \right\}\\
\dfrac{1}{{x + 1}} + \dfrac{2}{{x + 2}} \le \dfrac{3}{{x + 3}}\\
\to \dfrac{{\left( {x + 2} \right)\left( {x + 3} \right) + 2\left( {x + 1} \right)\left( {x + 3} \right) - 3\left( {x + 1} \right)\left( {x + 2} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)}} \le 0\\
\to \dfrac{{{x^2} + 5x + 6 + 2{x^2} + 8x + 6 - 3{x^2} - 9x - 9}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)}} \le 0\\
\to \dfrac{{4x + 3}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)}} \le 0
\end{array}\)
BXD:
x -∞ -3 -2 -1 -3/4 +∞
f(x) + // - // + // - 0 +
\(KL:x \in \left( { - 3; - 2} \right) \cup \left( { - 1; - \dfrac{3}{4}} \right]\)
\(\begin{array}{l}
23)DK:x \ne - 1\\
\dfrac{{\left( {2x - 1} \right){{\left( {x - 3} \right)}^4}}}{{ - x - 1}} \ge 0
\end{array}\)
BXD:
x -∞ -1 1/2 3(kép) +∞
f(x) - // + 0 - 0 -
\(KL:x \in \left( { - 1;\dfrac{1}{2}} \right] \cup \left\{ 3 \right\}\)
\(\begin{array}{l}
24)DK:x \ne \left\{ {2;7} \right\}\\
\dfrac{{{{\left( {x + 2} \right)}^4}\left( {x + 6} \right)}}{{{{\left( {x - 7} \right)}^3}{{\left( {x - 2} \right)}^2}}} \ge 0
\end{array}\)
BXD:
x -∞ -6 -2(kép) 2(kép) 7 +∞
f(x) + 0 - 0 - // - // +
\(KL:x \in \left( { - \infty ; - 6} \right] \cup \left( {7; + \infty } \right) \cup \left\{ { - 2} \right\}\)
