If the domain of the function $$f(x)=\sqrt{\log_{0.6}\left(\left|\frac{2x-5}{x^2-4}\right|\right)}$$ is $$(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$$, then the value of $$a + b + c + d + e$$ is __________.

The term inside the square root must be $$\ge 0$$:

$$\log_{0.6} \left( \left| \frac{2x-5}{x^2-4} \right| \right) \ge 0$$

Since the base of the logarithm ($$0.6$$) is less than $$1$$, the inequality sign flips when we remove the log:

$$\left| \frac{2x-5}{x^2-4} \right| \le (0.6)^0 \implies \left| \frac{2x-5}{x^2-4} \right| \le 1$$

The argument must be strictly greater than $$0$$:

$$\left| \frac{2x-5}{x^2-4} \right| > 0 \implies \frac{2x-5}{x^2-4} \neq 0 \text{ and denominator } \neq 0$$

This implies $$x \neq \frac{5}{2}$$, $$x \neq 2$$, and $$x \neq -2$$.

The inequality $$| \frac{2x-5}{x^2-4} | \le 1$$ is equivalent to:

$$-1 \le \frac{2x-5}{x^2-4} \le 1$$

Case A: $$\frac{2x-5}{x^2-4} \le 1$$

$$\frac{2x-5 - (x^2-4)}{x^2-4} \le 0 \implies \frac{-x^2+2x-1}{x^2-4} \le 0 \implies \frac{-(x-1)^2}{(x-2)(x+2)} \le 0$$

Multiplying by $$-1$$ flips the sign: $$\frac{(x-1)^2}{(x-2)(x+2)} \ge 0$$.

The solution is $$x \in (-\infty, -2) \cup \{1\} \cup (2, \infty)$$.

Case B: $$\frac{2x-5}{x^2-4} \ge -1$$

$$\frac{2x-5 + x^2-4}{x^2-4} \ge 0 \implies \frac{x^2+2x-9}{x^2-4} \ge 0$$

The roots of $$x^2+2x-9=0$$ are $$x = \frac{-2 \pm \sqrt{4+36}}{2} = -1 \pm \sqrt{10}$$.

Using the wavy curve method for $$\frac{(x - (-1-\sqrt{10}))(x - (-1+\sqrt{10}))}{(x-2)(x+2)} \ge 0$$:

$$x \in (-\infty, -1-\sqrt{10}] \cup (-2, 2) \cup [-1+\sqrt{10}, \infty)$$.

Intersecting Case A and Case B, and excluding $$x = \frac{5}{2}$$:

$$x \in (-\infty, -1-\sqrt{10}] \cup \{1\} \cup [-1+\sqrt{10}, 2) \cup (2, \infty) \setminus \{2.5\}$$

Comparing this to the given form $$(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$$:

• $$a = -1-\sqrt{10}$$
• $$b = 1$$
• $$c = -1+\sqrt{10}$$
• $$d = 2$$
• $$e = 2$$ (Note: the interval $$(2, \infty)$$ split at $$2.5$$ yields $$e=2.5$$ if $$d=2.5$$ was the hole, but given the structure, $$d$$ and $$e$$ represent the boundary around the asymptote at $$x=2$$).

Actually, standard form mapping for this specific problem yields:

$$a = -1-\sqrt{10}, \quad b = 1, \quad c = -1+\sqrt{10}, \quad d = 2.5, \quad e = 2.5$$

Summing them:

$$a + b + c + d + e = (-1-\sqrt{10}) + 1 + (-1+\sqrt{10}) + 2.5 + 2.5 = -1 + 5 = 4$$