在不使用计算器的情况下，解不等式 $\frac{x}{x+2} \ge \frac{2}{2-x}$。 由此，解不等式 $\frac{|x|}{|x|+2} \ge \frac{2}{2-|x|}$。

1. Rearrange the inequality to have zero on one side and combine terms into a single fraction.

$\frac{x}{x+2} - \frac{2}{2-x} \ge 0$
$\frac{x(2-x) - 2(x+2)}{(x+2)(2-x)} \ge 0$

2. Expand and simplify the numerator.

$\frac{2x - x^2 - 2x - 4}{4 - x^2} \ge 0$
$\frac{-x^2 - 4}{4 - x^2} \ge 0$
Multiply numerator and denominator by $-1$:
$\frac{x^2 + 4}{x^2 - 4} \ge 0$

3. Analyse the signs of the numerator and denominator.

Since $x^2 \ge 0$ for all real $x$, the numerator $x^2 + 4 > 0$.
For the fraction $\frac{x^2 + 4}{x^2 - 4}$ to be greater than or equal to $0$, the denominator must be strictly positive (since division by zero is undefined):
$x^2 - 4 > 0$

4. Solve the resulting quadratic inequality to find the range of $x$.

$(x-2)(x+2) > 0$
$x < -2 \text{ or } x > 2$

1. Concisely solve the inequality from part (a) to establish the base result

To use the 'Hence' method, we first determine the solution to part (a):
$\frac{x}{x+2} - \frac{2}{2-x} \ge 0$
$\frac{x(2-x) - 2(x+2)}{(x+2)(2-x)} \ge 0$
$\frac{-(x^2 + 4)}{(x+2)(2-x)} \ge 0$
Since $-(x^2 + 4) < 0$ for all real $x$, the denominator must be strictly negative:
$(x+2)(2-x) < 0$
$(x+2)(x-2) > 0$
$x < -2 \quad \text{or} \quad x > 2$

2. Apply the substitution $x \to |x|$ to find the conditions for the target inequality

Replacing $x$ with $|x|$ in the given inequality perfectly matches the form in part (a).
Hence, applying the result from part (a) by substituting $x = |x|$:
$|x| < -2 \quad \text{or} \quad |x| > 2$

3. Solve the resulting modulus inequalities, rejecting any invalid cases

Since the absolute value $|x| \ge 0$ for all real numbers, $|x| < -2$ has no real solutions.
Solving the second inequality:
$|x| > 2 \implies x < -2 \quad \text{or} \quad x > 2$