Đáp án:
$\begin{array}{l}+)\quad f(x) < 0 \Leftrightarrow x \in(-\sqrt2;\sqrt2)\backslash\{0\}\\ +) \quad f(x) >0\Leftrightarrow x \in (-\infty;-2)\cup (\sqrt2;+\infty)\end{array}$
Giải thích các bước giải:
$\begin{array}{l} \quad f(x) = \dfrac{x^2 + 4x + 4}{x^4 - 2x^2}\\ \to f(x) = \dfrac{(x+2)^2}{x^2(x-\sqrt2)(x+\sqrt2)}\\ \to f(x) = \left(\dfrac{x+2}{x}\right)^2\cdot\dfrac{1}{(x-\sqrt2)(x+\sqrt2)}\\ \text{Bảng xét dấu}\\ \begin{array}{|c|cr|} \hline x & -\infty&&-2&&-\sqrt2&&0&&\sqrt2&&+\infty\\ \hline \left(\dfrac{x+2}{x}\right)^2&&+&&+&&+&&+&&+&\\ \hline x - \sqrt2&&-&&-&&-&&-&&+&\\ \hline x + \sqrt2&&-&&-&&+&&+&&+&\\ \hline (x-\sqrt2)(x+\sqrt2)&&+&&+&&-&&-&&+&\\ \hline f(x)&&+&&+&&-&&-&&+&\\ \hline \end{array}\\ \text{Vậy:}\\ +)\quad f(x) < 0 \Leftrightarrow x \in(-\sqrt2;\sqrt2)\backslash\{0\}\\ +) \quad f(x) >0\Leftrightarrow x \in (-\infty;-2)\cup (\sqrt2;+\infty) \end{array}$
