解不等式 $\displaystyle \frac{x^{2} - ax - a}{x - a} \ge a$，其中 $a$ 为正实常数，答案用 $a$ 表示。 由此，通过使用合适的 $a$ 值，解不等式 $\displaystyle \frac{4e^{2x} - e^{x} - 1}{4e^{x} - 1} \ge \tfrac{1}{4}$ 答案用精确形式表示。

1. Rearrange the inequality so that one side is zero and express it as a single algebraic fraction.

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

2. Find the critical values by setting the numerator and denominator to zero.

For the denominator:
$x - a = 0 \implies x = a$

For the numerator:
$x^2 - 2ax + a^2 - a = 0$
By completing the square:
$(x - a)^2 - a = 0$
$(x - a)^2 = a$
Since $a$ is positive, we can take the square root of both sides:
$x - a = \pm \sqrt{a} \implies x = a \pm \sqrt{a}$

3. Determine the valid intervals using a sign line test for the critical values.

The critical values in increasing order are $a - \sqrt{a}$, $a$, and $a + \sqrt{a}$.
We test the sign of the fraction in each interval:

• For $x > a + \sqrt{a}$, the expression is $> 0$.
• For $a < x < a + \sqrt{a}$, the expression is $< 0$.
• For $a - \sqrt{a} < x < a$, the expression is $> 0$.
• For $x < a - \sqrt{a}$, the expression is $< 0$.

4. State the final solution, ensuring the denominator is not zero.

Since the inequality is $\ge 0$, we include the roots of the numerator ($x = a \pm \sqrt{a}$) but strictly exclude the root of the denominator ($x = a$).
$a - \sqrt{a} \le x < a \quad \text{or} \quad x \ge a + \sqrt{a}$

1. Solve the rational inequality from part (i) to establish the necessary foundational result.

$\frac{x^2 - ax - a}{x - a} - a \ge 0$
$\frac{x^2 - 2ax + a^2 - a}{x - a} \ge 0$
$\frac{(x - a)^2 - a}{x - a} \ge 0$
The roots of the numerator are given by $(x - a)^2 = a \implies x = a \pm \sqrt{a}$.
The root of the denominator is $x = a$.
Using a sign test with critical values $a - \sqrt{a}$, $a$, and $a + \sqrt{a}$ (where $a > 0$):
The solution to (i) is $a - \sqrt{a} \le x < a \text{ or } x \ge a + \sqrt{a}$.

2. Rewrite the target inequality to match the form of part (i) and identify the appropriate substitution.

Given:
$\frac{4e^{2x} - e^{x} - 1}{4e^{x} - 1} \ge \frac{1}{4}$
Divide the numerator and the denominator of the left-hand side by $4$:
$\frac{e^{2x} - \frac{1}{4}e^{x} - \frac{1}{4}}{e^{x} - \frac{1}{4}} \ge \frac{1}{4}$
Comparing this to $\frac{X^2 - aX - a}{X - a} \ge a$, we let $X = e^x$ and $a = \frac{1}{4}$.

3. Substitute $a = \frac{1}{4}$ and $X = e^x$ into the solution range obtained in part (i).

Substitute into $a - \sqrt{a} \le X < a \text{ or } X \ge a + \sqrt{a}$:
$\frac{1}{4} - \sqrt{\frac{1}{4}} \le e^x < \frac{1}{4} \quad \text{or} \quad e^x \ge \frac{1}{4} + \sqrt{\frac{1}{4}}$
$\frac{1}{4} - \frac{1}{2} \le e^x < \frac{1}{4} \quad \text{or} \quad e^x \ge \frac{1}{4} + \frac{1}{2}$
$-\frac{1}{4} \le e^x < \frac{1}{4} \quad \text{or} \quad e^x \ge \frac{3}{4}$

4. Solve for $x$, considering the domain of the exponential function.

Since $e^x > 0$ for all real $x$, the first inequality $-\frac{1}{4} \le e^x < \frac{1}{4}$ simplifies to $e^x < \frac{1}{4}$, which gives:
$e^x < \frac{1}{4} \implies x < \ln\left(\frac{1}{4}\right)$
For the second inequality:
$e^x \ge \frac{3}{4} \implies x \ge \ln\left(\frac{3}{4}\right)$