在不使用计算器的情况下，求不等式 $\frac{x^2 - x - 1}{x + 1} \le 1$ 的精确解。 由此，求不等式 $\frac{x^2 + |x| - 1}{1 - |x|} \le 1$ 的精确解。

1. Rearrange the inequality to compare a single rational fraction to zero.

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

For the numerator:
$x^2 - 2x - 2 = 0$
Using the quadratic formula:
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2}$
$x = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$

For the denominator:
$x + 1 = 0 \implies x = -1$

3. Determine the sign of the fraction in each interval defined by the critical values to find the solution.

The critical values in increasing order are $-1$, $1 - \sqrt{3} \approx -0.732$, and $1 + \sqrt{3} \approx 2.732$.

Testing the intervals:

• For $x < -1$ (e.g., $x = -2$): numerator $(+)$, denominator $(-)$ $\implies$ fraction is $(-)$
• For $-1 < x < 1 - \sqrt{3}$ (e.g., $x = -0.9$): numerator $(+)$, denominator $(+)$ $\implies$ fraction is $(+)$
• For $1 - \sqrt{3} < x < 1 + \sqrt{3}$ (e.g., $x = 0$): numerator $(-)$, denominator $(+)$ $\implies$ fraction is $(-)$
• For $x > 1 + \sqrt{3}$ (e.g., $x = 3$): numerator $(+)$, denominator $(+)$ $\implies$ fraction is $(+)$

We require the fraction to be $\le 0$. The roots $x = 1 \pm \sqrt{3}$ are included as they make the expression zero, but $x = -1$ is excluded as it makes the denominator undefined.
$x < -1 \text{ or } 1 - \sqrt{3} \le x \le 1 + \sqrt{3}$

1. State the solution to the inequality from part (a) concisely.

From part (a), we solve the inequality $\frac{u^2 - u - 1}{u + 1} \le 1$ by rearranging it to:
$\frac{u^2 - 2u - 2}{u + 1} \le 0$
Finding the critical values $u = -1$ and $u = 1 \pm \sqrt{3}$, we test the intervals to obtain the valid solution ranges:
$u < -1 \quad \text{or} \quad 1 - \sqrt{3} \le u \le 1 + \sqrt{3}$

2. Relate the inequality in part (b) to part (a) using a substitution.

Let $u = -|x|$. Substituting this into the algebraic expression from (a) gives:
$\frac{(-|x|)^2 - (-|x|) - 1}{(-|x|) + 1} = \frac{|x|^2 + |x| - 1}{1 - |x|} = \frac{x^2 + |x| - 1}{1 - |x|}$
This matches the inequality expression exactly. Therefore, solving (b) is equivalent to substituting $u = -|x|$ into the solution bounds from (a).

3. Solve the inequality for the first interval $u < -1$.

Substitute $u = -|x|$ into the first valid range:
$-|x| < -1$
$|x| > 1$
This yields the solution:
$x < -1 \quad \text{or} \quad x > 1$

4. Solve the inequality for the second interval $1 - \sqrt{3} \le u \le 1 + \sqrt{3}$.

Substitute $u = -|x|$ into the second valid range:
$1 - \sqrt{3} \le -|x| \le 1 + \sqrt{3}$
Multiplying the entire inequality by $-1$ reverses the inequality signs:
$-1 - \sqrt{3} \le |x| \le \sqrt{3} - 1$
Since the absolute value $|x| \ge 0$ for all real $x$, the left-hand condition $-1 - \sqrt{3} \le |x|$ is always satisfied. Thus, we only need to evaluate the right-hand condition:
$|x| \le \sqrt{3} - 1$
$-(\sqrt{3} - 1) \le x \le \sqrt{3} - 1$
$1 - \sqrt{3} \le x \le \sqrt{3} - 1$

5. Combine the valid ranges of $x$ for the final answer.

The complete set of solutions is the union of the valid ranges found from both substituted intervals:
$x < -1 \text{ or } 1 - \sqrt{3} \le x \le \sqrt{3} - 1 \text{ or } x > 1$