Đáp án:
\(x \in \left( { - 2;1} \right) \cup \left[ {1,337944335;2} \right) \cup \left( {3; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ { - 2;1;2;3} \right\}\\
\dfrac{1}{{{x^2} - 4}} \le \dfrac{{2x}}{{{x^2} - 4x + 3}}\\
\to \dfrac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)}} \le \dfrac{{2x}}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\\
\to \dfrac{{2x\left( {{x^2} - 4} \right) - {x^2} + 4x - 3}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 1} \right)\left( {x - 3} \right)}} \ge 0\\
\to \dfrac{{2{x^3} - 4x - {x^2} + 4x - 3}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 1} \right)\left( {x - 3} \right)}} \ge 0\\
\to \dfrac{{2{x^3} - {x^2} - 3}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 1} \right)\left( {x - 3} \right)}} \ge 0\\
Xet:2{x^3} - {x^2} - 3 = 0\\
\to x = 1,337944335
\end{array}\)
BXD:
x -∞ -2 1 1,33 2 3 +∞
f(x) - // + // - 0 + // - // +
\(KL:x \in \left( { - 2;1} \right) \cup \left[ {1,337944335;2} \right) \cup \left( {3; + \infty } \right)\)
