求不等式 $\frac{2x+1}{x^2-1} \le 1$ 的精确解。 曲线 $G$ 的方程为 $ay^2 + bx^3 + cx = 2$，其中 $a$、$b$ 和 $c$ 是常数。已知 $G$ 经过点 $(1, \sqrt{3})$ 并且在点 $(-1, 1)$ 处的斜率为 $-\frac{3}{2}$。求曲线 $G$ 的方程。

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

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

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

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

For the denominator:
$x^2 - 1 = 0 \implies x = \pm 1$

The critical values are $x = -1, 1-\sqrt{3}, 1, 1+\sqrt{3}$.

3. Determine the sign of the rational expression in the intervals defined by the critical values.

Arrange the critical values in increasing order: $-1 < 1-\sqrt{3} < 1 < 1+\sqrt{3}$.

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

• $x < -1$: Positive $(>0)$
• $-1 < x < 1-\sqrt{3}$: Negative $(<0)$
• $1-\sqrt{3} < x < 1$: Positive $(>0)$
• $1 < x < 1+\sqrt{3}$: Negative $(<0)$
• $x > 1+\sqrt{3}$: Positive $(>0)$

The inequality requires the expression to be $\ge 0$. The roots of the numerator ($x = 1 \pm \sqrt{3}$) are included in the solution, while the roots of the denominator ($x = \pm 1$) are excluded.
$x < -1 \text{ or } 1-\sqrt{3} \le x < 1 \text{ or } x \ge 1+\sqrt{3}$

1. Form an equation using the point $(1, \sqrt{3})$.

Since the curve passes through $(1, \sqrt{3})$, substitute $x = 1$ and $y = \sqrt{3}$ into the equation of the curve $ay^2 + bx^3 + cx = 2$:
$a(\sqrt{3})^2 + b(1)^3 + c(1) = 2$
$3a + b + c = 2 \quad \text{--- (1)}$

2. Form a second equation using the point $(-1, 1)$.

The gradient is given at $(-1, 1)$, which means the point $(-1, 1)$ also lies on the curve. Substitute $x = -1$ and $y = 1$:
$a(1)^2 + b(-1)^3 + c(-1) = 2$
$a - b - c = 2 \quad \text{--- (2)}$

3. Find an expression involving the gradient using implicit differentiation.

Differentiate $ay^2 + bx^3 + cx = 2$ implicitly with respect to $x$:
$2ay \frac{dy}{dx} + 3bx^2 + c = 0$

4. Form a third equation using the given gradient.

Substitute $x = -1$, $y = 1$, and $\frac{dy}{dx} = -\frac{3}{2}$ into the differentiated equation:
$2a(1)\left(-\frac{3}{2}\right) + 3b(-1)^2 + c = 0$
$-3a + 3b + c = 0 \quad \text{--- (3)}$

5. Solve the system of three linear equations to find $a$, $b$, and $c$.

Adding equation (1) and equation (2):
$(3a + b + c) + (a - b - c) = 2 + 2$
$4a = 4 \implies a = 1$
Substitute $a = 1$ into equation (2):
$1 - b - c = 2 \implies b + c = -1 \implies c = -1 - b$
Substitute $a = 1$ and $c = -1 - b$ into equation (3):
$-3(1) + 3b + (-1 - b) = 0$
$-4 + 2b = 0 \implies 2b = 4 \implies b = 2$
Find $c$:
$c = -1 - 2 = -3$
The constants are $a = 1$, $b = 2$, and $c = -3$.

6. State the final equation of the curve.

Substituting $a=1$, $b=2$, and $c=-3$ back into the given form, the equation of the curve is:
$y^2 + 2x^3 - 3x = 2$