不使用计算器，求解

1. Bring all terms to one side and combine them into a single rational expression.

$1 + \frac{6}{x} - x \ge 0$
$\frac{x + 6 - x^2}{x} \ge 0$

2. Factorise the numerator and identify the critical values.

$\frac{-(x^2 - x - 6)}{x} \ge 0$
$\frac{(x-3)(x+2)}{x} \le 0$
Critical values are $x = 3, -2, 0$.

3. Perform a sign test to determine the valid intervals, noting the denominator restriction $x \neq 0$.

Testing intervals around the critical values for $\frac{(x-3)(x+2)}{x} \le 0$:

• For $x < -2$, expression is $\frac{(-)(-)}{(-)} \rightarrow \text{negative} \le 0$ (Valid)
• For $-2 < x < 0$, expression is $\frac{(-)(+)}{(-)} \rightarrow \text{positive} > 0$ (Invalid)
• For $0 < x < 3$, expression is $\frac{(-)(+)}{(+)} \rightarrow \text{negative} \le 0$ (Valid)
• For $x > 3$, expression is $\frac{(+)(+)}{(+)} \rightarrow \text{positive} > 0$ (Invalid)

Therefore, the solution is $x \le -2 \text{ or } 0 < x \le 3$.

1. Solve the inequality from part (i) to use as a prerequisite

To solve part (ii) efficiently, we first establish the solution to part (i):
$1 + \frac{6}{x} - x \ge 0$
$\frac{x + 6 - x^2}{x} \ge 0$
$\frac{x^2 - x - 6}{x} \le 0$
$\frac{(x-3)(x+2)}{x} \le 0$
Using a sign test with critical values $x = -2, 0, 3$, the solution to part (i) is:
$x \le -2 \quad \text{or} \quad 0 < x \le 3$

2. Substitute $x$ with $|x|$ based on the structure of the new inequality

The inequality in part (ii) is $1 + \frac{6}{|x|} \ge |x|$.
This is identical to the inequality in (i) with $x$ replaced by $|x|$.
Substituting $x = |x|$ into the solution from part (i) gives:
$|x| \le -2 \quad \text{or} \quad 0 < |x| \le 3$

3. Solve the resulting modulus inequalities

Analyse the two conditions separately:

1. $|x| \le -2$ has no real solution because $|x| \ge 0$ for all real $x$.
2. For the second condition $0 < |x| \le 3$:
   • $|x| > 0 \implies x \neq 0$
   • $|x| \le 3 \implies -3 \le x \le 3$

Combining these constraints, we exclude $x = 0$ from the interval $[-3, 3]$:
$-3 \le x < 0 \quad \text{or} \quad 0 < x \le 3$