Đáp án:
\(x \in \left( { - \infty ; - 2} \right) \cup \left( {\dfrac{{5 - \sqrt {17} }}{2};2} \right) \cup \left( {2;\dfrac{{5 + \sqrt {17} }}{2}} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \pm 2\\
\dfrac{1}{{x + 2}} < \dfrac{1}{{{{\left( {x - 2} \right)}^2}}}\\
\to \dfrac{{{x^2} - 4x + 4 - x - 2}}{{\left( {x + 2} \right){{\left( {x - 2} \right)}^2}}} < 0\\
\to \dfrac{{{x^2} - 5x + 2}}{{\left( {x + 2} \right){{\left( {x - 2} \right)}^2}}} < 0\\
Xét:\dfrac{{{x^2} - 5x + 2}}{{\left( {x + 2} \right){{\left( {x - 2} \right)}^2}}} = 0\\
\to {x^2} - 5x + 2 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{5 + \sqrt {17} }}{2}\\
x = \dfrac{{5 - \sqrt {17} }}{2}
\end{array} \right.
\end{array}\)
BXD:
x -∞ -2 \(\dfrac{{5 - \sqrt {17} }}{2}\) 2(kép) \(\dfrac{{5 - \sqrt {17} }}{2}\) +∞
f(x) - // + 0 - // - 0 +
\(KL:x \in \left( { - \infty ; - 2} \right) \cup \left( {\dfrac{{5 - \sqrt {17} }}{2};2} \right) \cup \left( {2;\dfrac{{5 + \sqrt {17} }}{2}} \right)\)
