在不使用计算器的情况下，解不等式 $\frac{(x^2 - 2x + 4)(x - 3)}{(x + 2)} \ge 0.$ 由此解不等式 $\frac{(x^2 - 2|x| + 4)(|x| - 3)}{(|x| + 2)} \ge 0.$

1. Show that the quadratic factor in the numerator is always positive.

By completing the square for $x^2 - 2x + 4$:
$x^2 - 2x + 4 = (x - 1)^2 - 1^2 + 4$
$x^2 - 2x + 4 = (x - 1)^2 + 3$
Since $(x - 1)^2 \ge 0$ for all real $x$, it follows that $(x - 1)^2 + 3 \ge 3 > 0$.
Thus, $x^2 - 2x + 4$ is always strictly positive.

2. Simplify the inequality by removing the positive factor.

Since $x^2 - 2x + 4 > 0$ for all real $x$, dividing both sides of the inequality by this expression will not change the inequality sign.
$\frac{x - 3}{x + 2} \ge 0$

3. Convert the rational inequality to a polynomial inequality.

Multiply both sides by $(x + 2)^2$. Since $(x + 2)^2 > 0$ for $x \neq -2$, the inequality sign remains unchanged:
$\frac{x - 3}{x + 2} \times (x + 2)^2 \ge 0 \times (x + 2)^2$
$(x - 3)(x + 2) \ge 0$
Note that $x \neq -2$ to avoid a zero denominator in the original rational expression.

4. Solve the resulting quadratic inequality and deduce the final range.

The critical values for $(x - 3)(x + 2) \ge 0$ are $x = -2$ and $x = 3$.
Because it is a '$\ge 0$' inequality with a positive $x^2$ coefficient, the solution lies outside the roots:
$x \le -2 \text{ or } x \ge 3$
However, because $x \neq -2$ (undefined denominator), we reject $x = -2$.
Therefore, the solution is:
$x < -2 \text{ or } x \ge 3$

1. Concisely solve the inequality from part (i) to find the base solution range.

For the quadratic factor, $x^2 - 2x + 4 = (x - 1)^2 + 3 > 0$ for all real $x$.
Dividing both sides by this strictly positive factor, the inequality in (i) simplifies to:
$\frac{x - 3}{x + 2} \ge 0$
By testing intervals around the critical values $x = -2$ (where the expression is undefined) and $x = 3$ (where it is zero), the solution to (i) is:
$x < -2 \quad \text{or} \quad x \ge 3$

2. Substitute $|x|$ for $x$ in the base solution, recognising that $x^2 = |x|^2$.

The inequality in (ii) can be rewritten as:
$\frac{(|x|^2 - 2|x| + 4)(|x| - 3)}{|x| + 2} \ge 0$
This is the exact same inequality as in (i), but with $x$ replaced by $|x|$. Using the result from (i), we substitute $x$ with $|x|$ to get:
$|x| < -2 \quad \text{or} \quad |x| \ge 3$

3. Solve the resulting modulus inequalities for $x$.

Since the absolute value of any real number is non-negative ($|x| \ge 0$), the inequality $|x| < -2$ has no real solution.
For the remaining inequality:
$|x| \ge 3$
This gives the solution:
$x \le -3 \quad \text{or} \quad x \ge 3$