$\left | \frac{x^{2}-2}{x} \right| \cdot\left | \frac{x}{x^{2}-2} \right|\neq 1$

Question

I'm sorry if I ask something trivial and about basic math. Why I cannot simplify two absolute values?

I'm considering

$$\frac{\partial}{\partial x}\left [ \log\left | \frac{x^{2}-2}{x} \right| \right ]=\left | \frac{x^{2}-2}{x} \right| \cdot\left | \frac{x}{x^{2}-2} \right|\cdot \frac{x}{x^{2}-2}\cdot\frac{2x^{2}-2-x^{2}}{x^{2}}\mapsto \frac{x^{2}+2}{x^{3}-2x}.$$

Why the following equality is true? $$\left | \frac{x^{2}-2}{x} \right| \cdot\left | \frac{x}{x^{2}-2} \right|\neq 1.$$

Answer 1

For all $a,b \in \mathbb{R}$ it is true that $$\vert a\cdot b\vert = \vert a \vert \cdot \vert b \vert \tag{1}$$

This is the multiplicative property. As stated in the comments by Wore and Zestylemonzi you'll have that, for $x \in \mathbb{R}-\{0\}$ we have that

$$\frac{x^2-2}{x} \in \mathbb{R}$$

and for $x \in \mathbb{R}-\{-\sqrt{2},\sqrt{2}\}$ you'll get

$$\frac{x}{x^2-2}\in \mathbb{R}$$

so using $(1)$ you get that the equality is true if $x$ satisfies both $x \neq 0$ and $x\neq \sqrt{2}$ or $-\sqrt{2}$.