不使用计算器，解不等式 $\dfrac{11x-19}{x^2+2x-8} \le 2$。

1. Solve the rational inequality from the previous part

Let $u$ be the variable in the original inequality.
$\frac{11u-19}{u^2+2u-8} - 2 \le 0$
$\frac{11u-19-2(u^2+2u-8)}{(u+4)(u-2)} \le 0$
$\frac{-2u^2+7u-3}{(u+4)(u-2)} \le 0$
$\frac{2u^2-7u+3}{(u+4)(u-2)} \ge 0$
$\frac{(2u-1)(u-3)}{(u+4)(u-2)} \ge 0$
Finding the critical values $u = -4, \frac{1}{2}, 2, 3$ and testing intervals gives:
$u < -4 \quad \text{or} \quad \frac{1}{2} \le u < 2 \quad \text{or} \quad u \ge 3$

2. Apply the substitution $u = e^{-x}$ and discard the invalid range

Substitute $u = e^{-x}$ into the solution ranges:
$e^{-x} < -4 \quad \text{or} \quad \frac{1}{2} \le e^{-x} < 2 \quad \text{or} \quad e^{-x} \ge 3$
Since $e^{-x} > 0$ for all real $x$, the inequality $e^{-x} < -4$ has no real solution.

3. Solve the remaining exponential inequalities exactly

For $\frac{1}{2} \le e^{-x} < 2$:
Take the natural logarithm on all parts:
$\ln\left(\frac{1}{2}\right) \le -x < \ln 2$
$-\ln 2 \le -x < \ln 2$
Multiply by $-1$ and reverse the inequality signs:
$-\ln 2 < x \le \ln 2$

For $e^{-x} \ge 3$:
$-x \ge \ln 3$
$x \le -\ln 3$

Combining the valid intervals gives the final exact solution.