本题不得使用计算器求解。

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

$\frac{x-6}{4x^2+x-5} - 1 \ge 0$
$\frac{x-6 - (4x^2+x-5)}{4x^2+x-5} \ge 0$
$\frac{-4x^2-1}{4x^2+x-5} \ge 0$

2. Factorise the denominator and simplify the inequality signs.

$\frac{-(4x^2+1)}{(4x+5)(x-1)} \ge 0$
$\frac{4x^2+1}{(4x+5)(x-1)} \le 0$

3. Use sign analysis to solve the rational inequality.

Since $x^2 \ge 0$ for all real $x$, we have $4x^2+1 > 0$.

For the rational expression to be $\le 0$, the denominator must be strictly negative (as division by zero is undefined):
$(4x+5)(x-1) < 0$
$-\frac{5}{4} < x < 1$

1. Solve the inequality from part (i) as its solution is required to evaluate part (ii).

Since part (ii) uses 'Hence', we first establish the solution to part (i):
$\frac{x-6}{4x^2+x-5} \ge 1$
$\frac{x-6-(4x^2+x-5)}{4x^2+x-5} \ge 0$
$\frac{-4x^2-1}{4x^2+x-5} \ge 0$
Since $-4x^2-1 = -(4x^2+1) < 0$ for all real $x$, the denominator must be strictly negative for the fraction to be non-negative:
$4x^2+x-5 < 0$
$(4x+5)(x-1) < 0$
This gives the solution for part (i) as:
$-\frac{5}{4} < x < 1$

2. Show that the inequality in part (ii) can be obtained by an algebraic substitution into the expression from part (i).

Observe the relationship between the expression in part (ii) and part (i). Substitute $X = \frac{1}{x}$ into the expression from part (i):
$\frac{\frac{1}{x}-6}{4\left(\frac{1}{x}\right)^2+\frac{1}{x}-5} = \frac{\frac{1-6x}{x}}{\frac{4+x-5x^2}{x^2}} = \frac{x(1-6x)}{4+x-5x^2} = \frac{x-6x^2}{4+x-5x^2}$

3. Use the 'Hence' dependency to set up a new inequality based on the solution to part (i).

Because the expressions are equivalent under the substitution $X = \frac{1}{x}$, the inequality $\frac{x-6x^2}{4+x-5x^2} \ge 1$ holds when the substituted variable falls within the valid range found in part (i):
$-\frac{5}{4} < \frac{1}{x} < 1$

4. Solve the compound inequality analytically.

Split the compound inequality into two separate parts and solve algebraically.

For $\frac{1}{x} < 1$:
$\frac{1-x}{x} < 0$
This yields $x < 0$ or $x > 1$.

For $\frac{1}{x} > -\frac{5}{4}$:
$\frac{1}{x} + \frac{5}{4} > 0 \implies \frac{4+5x}{4x} > 0$
This yields $x < -\frac{4}{5}$ or $x > 0$.

5. Find the intersection of the two solution sets to determine the final range.

Take the intersection of ($x < 0$ or $x > 1$) and ($x < -\frac{4}{5}$ or $x > 0$):

• For negative values, $x$ must satisfy $x < 0$ AND $x < -\frac{4}{5}$, which simplifies to $x < -\frac{4}{5}$.
• For positive values, $x$ must satisfy $x > 1$ AND $x > 0$, which simplifies to $x > 1$.

Thus, the final combined solution is:
$x < -\frac{4}{5} \text{ or } x > 1$