Đáp án:
1) \(x \in \left( { - \infty ;\dfrac{{ - 1 - \sqrt 5 }}{2}} \right] \cup \left( { - 1;\dfrac{{ - 1 + \sqrt 5 }}{2}} \right] \cup \left( {0;1} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne \left\{ { - 1;0;1} \right\}\\
\dfrac{2}{x} + \dfrac{1}{{x - 1}} - \dfrac{1}{{x + 1}} \le 0\\
\to \dfrac{{2{x^2} - 2 + x\left( {x + 1} \right) - x\left( {x - 1} \right)}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}} \le 0\\
\to \dfrac{{2{x^2} - 2 + {x^2} + x - {x^2} + x}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}} \le 0\\
\to \dfrac{{2{x^2} + 2x - 2}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}} \le 0\\
Xét:2{x^2} + 2x - 2 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 1 + \sqrt 5 }}{2}\left( {N1} \right)\\
x = \dfrac{{ - 1 - \sqrt 5 }}{2}\left( {N2} \right)
\end{array} \right.
\end{array}\)
BXD:
x -∞ N2 -1 N1 0 1 +∞
f(x) - 0 + // - 0 + // - // +
\(KL:x \in \left( { - \infty ;\dfrac{{ - 1 - \sqrt 5 }}{2}} \right] \cup \left( { - 1;\dfrac{{ - 1 + \sqrt 5 }}{2}} \right] \cup \left( {0;1} \right)\)
\(\begin{array}{l}
2)DK:x \ne \left\{ { - 3; - 2;0} \right\}\\
\dfrac{{{x^2} - 5x + 6}}{{{x^2} + 5x + 6}} \ge \dfrac{{x + 1}}{x}\\
\to \dfrac{{{x^3} - 5{x^2} + 6x - {x^3} - 5{x^2} - 6x - {x^2} - 5x - 6}}{{x\left( {x + 2} \right)\left( {x + 3} \right)}} \ge 0\\
\to \dfrac{{ - 11{x^2} - 5x - 6}}{{x\left( {x + 2} \right)\left( {x + 3} \right)}} \ge 0\\
\to x\left( {x + 2} \right)\left( {x + 3} \right) < 0\left( {do: - 11{x^2} - 5x - 6 < 0\forall x \ne \left\{ { - 3; - 2;0} \right\}} \right)
\end{array}\)
BXD:
x -∞ -3 -2 0 +∞
f(x) - // + // - // +
\(KL:x \in \left( { - \infty ; - 3} \right) \cup \left( { - 2;0} \right)\)
