已知 $a \in \mathbb{R}$ 且 $a > 1$，利用代数方法求解不等式 $\frac{ax - a + 2}{(x - a)(1 - x)} \le 1 .$ 由此求出满足 $\frac{a|x| - a + 2}{(|x| - a)(1 - |x|)} \le 1$ 的 $x$ 的取值集合。

1. Rearrange the inequality to have 0 on one side and express it as a single algebraic fraction.

$\frac{ax - a + 2}{(x - a)(1 - x)} - 1 \le 0$

$\frac{ax - a + 2 - (x - a)(1 - x)}{(x - a)(1 - x)} \le 0$

$\frac{ax - a + 2 - (x - x^2 - a + ax)}{(x - a)(1 - x)} \le 0$

$\frac{x^2 - x + 2}{(x - a)(1 - x)} \le 0$

2. Analyse the sign of the numerator.

Consider the numerator $x^2 - x + 2$:
Completing the square, we get:
$x^2 - x + 2 = \left(x - \frac{1}{2}\right)^2 + \frac{7}{4}$
Since $\left(x - \frac{1}{2}\right)^2 \ge 0$ for all real $x$, it follows that $x^2 - x + 2 > 0$ for all real $x$.
(Alternatively, the discriminant $\Delta = (-1)^2 - 4(1)(2) = -7 < 0$, and since the coefficient of $x^2$ is positive, the quadratic is strictly positive.)

3. Deduce the required inequality for the denominator and solve it.

Since the numerator is always strictly positive, for the whole fraction to be $\le 0$, the denominator must be strictly negative (note that the denominator cannot be equal to $0$ as division by zero is undefined).
$(x - a)(1 - x) < 0$
Multiplying both sides by $-1$ reverses the inequality sign:
$(x - a)(x - 1) > 0$
Given that $a > 1$, the critical values are $1$ and $a$, and they are ordered $1 < a$. Solving this quadratic inequality yields:
$x < 1 \text{ or } x > a$

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

$\frac{ax - a + 2}{(x - a)(1 - x)} - 1 \le 0$\n$\frac{ax - a + 2 - (-x^2 + x + ax - a)}{(x - a)(1 - x)} \le 0$\n$\frac{x^2 - x + 2}{(x - a)(1 - x)} \le 0$\nSince $x^2 - x + 2 = \left(x - \frac{1}{2}\right)^2 + \frac{7}{4} > 0$ for all real $x$, we require the denominator to be negative:\n$(x - a)(1 - x) < 0 \implies (x - a)(x - 1) > 0$\nGiven $a > 1$, the solution to part (i) is $x < 1 \text{ or } x > a$.

2. Apply the substitution $x \to |x|$ to the result of part (i).

Comparing the target inequality with the one in part (i), we can substitute $x$ with $|x|$ in the solution set:\n$|x| < 1 \quad \text{or} \quad |x| > a$

3. Solve the resulting modulus inequalities to find the final range of $x$.

For $|x| < 1$:\n$-1 < x < 1$\nFor $|x| > a$, since we are given $a > 1 > 0$:\n$x < -a \quad \text{or} \quad x > a$\nCombining these disjoint intervals, the set of values of $x$ is:\n$x < -a \quad \text{or} \quad -1 < x < 1 \quad \text{or} \quad x > a$