解答本题时请勿使用计算器。

1. Rearrange the inequality to have zero on one side and combine the terms into a single algebraic fraction.

$\frac{6x^2 + 2x - 3}{2x - 1} - 2(x + 1) \ge 0$

$\frac{6x^2 + 2x - 3 - 2(x + 1)(2x - 1)}{2x - 1} \ge 0$

2. Expand and simplify the numerator.

$2(x + 1)(2x - 1) = 2(2x^2 + x - 1) = 4x^2 + 2x - 2$

$\frac{6x^2 + 2x - 3 - (4x^2 + 2x - 2)}{2x - 1} \ge 0$

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

3. Identify the critical values by setting the numerator and denominator to zero.

Numerator: $2x^2 - 1 = 0 \implies x^2 = \frac{1}{2} \implies x = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2}$

Denominator: $2x - 1 = 0 \implies x = \frac{1}{2}$

4. Use a sign test to determine the intervals satisfying the inequality.

The critical values in ascending order are $-\frac{\sqrt{2}}{2}$, $\frac{1}{2}$, and $\frac{\sqrt{2}}{2}$ (since $\frac{\sqrt{2}}{2} \approx 0.707 > 0.5$).

Testing the sign of $\frac{2x^2 - 1}{2x - 1}$ in each interval:

• $x < -\frac{\sqrt{2}}{2}$ (e.g., $x = -1$): $\frac{2 - 1}{-2 - 1} = -\frac{1}{3} < 0$
• $-\frac{\sqrt{2}}{2} < x < \frac{1}{2}$ (e.g., $x = 0$): $\frac{-1}{-1} = 1 > 0$
• $\frac{1}{2} < x < \frac{\sqrt{2}}{2}$ (e.g., $x = 0.6$): $\frac{2(0.36) - 1}{1.2 - 1} = \frac{-0.28}{0.2} < 0$
• $x > \frac{\sqrt{2}}{2}$ (e.g., $x = 1$): $\frac{2 - 1}{2 - 1} = 1 > 0$

The fraction must be $\ge 0$. The roots of the numerator are included, but the root of the denominator ($x = \frac{1}{2}$) must be excluded to avoid division by zero.

Thus, $-\frac{\sqrt{2}}{2} \le x < \frac{1}{2}$ or $x \ge \frac{\sqrt{2}}{2}$.

1. Solve the inequality from part [0] to establish the foundation for substitution.

$\frac{6x^2 + 2x - 3}{2x - 1} - 2(x + 1) \ge 0$
$\frac{6x^2 + 2x - 3 - 2(2x^2 + x - 1)}{2x - 1} \ge 0$
$\frac{2x^2 - 1}{2x - 1} \ge 0$
Finding the critical values, we set the numerator and denominator to zero:
$2x^2 - 1 = 0 \implies x = \pm \frac{1}{\sqrt{2}}$
$2x - 1 = 0 \implies x = \frac{1}{2}$
Analysing the signs of the expression across these intervals gives the solution for part [0]:
$-\frac{1}{\sqrt{2}} \le x < \frac{1}{2} \quad \text{or} \quad x \ge \frac{1}{\sqrt{2}}$

2. Substitute $x$ with $|x|$ to relate the target inequality to the previous part.

Since $x^2 = |x|^2$, the target inequality is algebraically identical to the one in part [0] but with $|x|$ replacing $x$:
$\frac{6|x|^2 + 2|x| - 3}{2|x| - 1} \ge 2(|x| + 1)$
Substituting $|x|$ into the solution from part [0]:
$-\frac{1}{\sqrt{2}} \le |x| < \frac{1}{2} \quad \text{or} \quad |x| \ge \frac{1}{\sqrt{2}}$

3. Solve the first inequality involving the modulus.

For the range $-\frac{1}{\sqrt{2}} \le |x| < \frac{1}{2}$, we note that $|x| \ge 0$ is naturally true for all real values of $x$. Thus, we only need to solve $|x| < \frac{1}{2}$:
$-\frac{1}{2} < x < \frac{1}{2}$

4. Solve the second inequality involving the modulus and combine the valid ranges.

For the range $|x| \ge \frac{1}{\sqrt{2}}$, expanding the modulus condition gives:
$x \le -\frac{1}{\sqrt{2}} \quad \text{or} \quad x \ge \frac{1}{\sqrt{2}}$
Combining all the separate valid ranges of $x$ gives the final continuous solution set.