Giải các bất phương trình sau
1) 2. 1. 1
__ + _____ – _____ < bằng 0.
X. X-1 X+1.
2) x2 -5x+6 x+1
_________ > bằng. _____
X2+5x+6. X
Giúp em với ạ
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) – // + // – // +
\(KL:x \in \left( { – \infty ; – 3} \right) \cup \left( { – 2;0} \right)\)