将 $\dfrac{33-8x}{x^2+2x-15}+2$ 化简为单个代数分式。由此，在不使用计算器的前提下，求不等式 $\dfrac{33-8x}{x^2+2x-15}>-2$ 的精确解。 利用你在 (a) 部分的解答，求满足 $\dfrac{33-8e^{2x}}{e^{4x}+2e^{2x}-15}>-2$ 的 $x$ 的取值集合。

1. Combine the expression into a single algebraic fraction.

$\frac{33-8x}{x^2+2x-15}+2$
$= \frac{33-8x+2(x^2+2x-15)}{x^2+2x-15}$
$= \frac{33-8x+2x^2+4x-30}{x^2+2x-15}$
$= \frac{2x^2-4x+3}{x^2+2x-15}$

2. Rewrite the inequality to compare with zero and analyse the sign of the numerator.

$\frac{33-8x}{x^2+2x-15} > -2$
$\frac{33-8x}{x^2+2x-15} + 2 > 0$
$\frac{2x^2-4x+3}{x^2+2x-15} > 0$
Check the discriminant ($\Delta$) of the numerator $2x^2-4x+3$:
$\Delta = (-4)^2 - 4(2)(3) = 16 - 24 = -8$
Since the discriminant is negative ($-8 < 0$) and the coefficient of $x^2$ is positive ($2 > 0$), the quadratic expression is always positive, meaning $2x^2-4x+3 > 0$ for all real $x$.

3. Determine the required condition for the denominator and solve the resulting inequality.

For the rational fraction to be positive when the numerator is strictly positive, the denominator must also be strictly positive:
$x^2+2x-15 > 0$
Factorising the quadratic expression:
$(x+5)(x-3) > 0$
Solving this inequality gives:
$x < -5 \text{ or } x > 3$

1. Apply the result from part (a) by using the substitution $x = e^{2x}$.

From part (a), solving $\dfrac{33-8x}{x^2+2x-15} > -2$ yields the solution $x < -5$ or $x > 3$.

Replacing $x$ with $e^{2x}$ in the inequality gives:
$e^{2x} < -5 \quad \text{or} \quad e^{2x} > 3$

2. Solve the resulting exponential inequalities and reject any invalid cases.

Since the exponential function $e^{2x} > 0$ for all real $x$, the inequality $e^{2x} < -5$ has no real solution.

Solving the valid inequality:
$e^{2x} > 3$
Taking the natural logarithm on both sides:
$2x > \ln 3$
$x > \frac{1}{2}\ln 3$