11.29 Find the rational roots of the equation

$$
\frac{\sqrt{x+2}}{|x|}+\frac{|x|}{\sqrt{x+2}}=\frac{4}{3} \sqrt{3}
$$

Solve the inequalities (11.30-11.31):
11.29 We find the domain of the equation: $x+2>0$, hence $x>-2$, and $x \neq 0$, i.e., $(-2,0) \cup(0, \infty)$. Setting $\frac{\sqrt{x+2}}{|x|}=y$, we get the equation

$$
y+\frac{1}{y}=\frac{4 \sqrt{3}}{3}, \text { or } 3 y^{2}-4 \sqrt{3} y+3=0
$$

From this, $\quad y_{1,2}=\frac{2 \sqrt{3} \pm \sqrt{12-9}}{3}=\frac{2 \sqrt{3} \pm \sqrt{3}}{3} ; \quad y_{1}=\sqrt{3}$,
$y_{2}=\frac{\sqrt{3}}{3} \cdot$ Next, we solve two equations:

1) $\frac{\sqrt{x+2}}{|x|}=\sqrt{3} ; \frac{x+2}{x^{2}}=3 ; 3 x^{2}-x-2=0$ ；

$x_{1}=-\frac{2}{3}, x_{2}=1$.

These are rational roots and they lie within the domain of the equation.
2) $\frac{\sqrt{x+2}}{|x|}=\frac{\sqrt{3}}{3}$

$$
\frac{x+2}{x^{2}}=\frac{1}{3} ; \quad x^{2}-3 x-6=0
$$

$x_{1,2}=\frac{3 \pm \sqrt{9+24}}{2}$.

These are irrational roots, so they do not satisfy the condition of the problem.

Answer: $x_{1}=-\frac{2}{3}, x_{2}=1$.