The domain of $f\left(x\right)=\sqrt{{\log }_{0.6}\left(\frac{\left|2x-5\right|}{\left|x^2-4\right|}\right)}$ is

To find the domain of the function given by:
$f\left(x\right)=\sqrt{{\log }_{0.6}\left(\frac{\left|2x-5\right|}{\left|x^2-4\right|}\right)}$
we must identify the conditions for which the expression under the square root is real and well-defined.

Step 1: Conditions for the logarithm and denominator
First, the denominator of the fraction inside the absolute values must not be zero:
$\left|x^2-4\right|\ne 0\Rightarrow x\ne \pm 2$
Second, the argument of the logarithm must be strictly positive. Since it is a ratio of absolute values, it is always non-negative where defined. For it to be strictly positive, the numerator must not be zero:
$\left|2x-5\right|\ne 0\Rightarrow x\ne \frac{5}{2}$

Step 2: Condition for the square root
For the square root to be defined, its operand must be non-negative:
${\log }_{0.6}\left(\frac{\left|2x-5\right|}{\left|x^2-4\right|}\right)\ge 0$
Since the base of the logarithm is $0.6$, which is strictly between 0 and 1, the inequality reverses when we remove the logarithm:
$\frac{\left|2x-5\right|}{\left|x^2-4\right|}\le {0.6}^0\Rightarrow \frac{\left|2x-5\right|}{\left|x^2-4\right|}\le 1$
Since $\left|x^2-4\right|>0$ for $x\ne \pm 2$, we can multiply both sides by it:
$\left|2x-5\right|\le \left|x^2-4\right|$

Step 3: Solving the inequality
Squaring both sides of the inequality, we get:
${\left(2x-5\right)}^2\le {\left(x^2-4\right)}^2$
Rearranging the terms:
${\left(x^2-4\right)}^2-{\left(2x-5\right)}^2\ge 0$
Using the difference of squares identity $A^2-B^2=(A-B)(A+B)$:
$\left(x^2-4-(2x-5)\right)\left(x^2-4+2x-5\right)\ge 0$
Simplifying the terms inside the parentheses:
$\left(x^2-2x+1\right)\left(x^2+2x-9\right)\ge 0$
Notice that $x^2-2x+1={(x-1)}^2$. The inequality becomes:
${\left(x-1\right)}^2\left(x^2+2x-9\right)\ge 0$

Step 4: Finding the roots of the quadratic
The term ${(x-1)}^2$ is always non-negative. For the product to be non-negative, either $x=1$ or we must have:
$x^2\ge 0$
Let us find the roots of $x^2+2x-9=0$ using the quadratic formula:
$x=\frac{-2\pm \sqrt{2^2-4\left(1\right)(-9)}}{2}=\frac{-2\pm \sqrt{40}}{2}=-1\pm \sqrt{10}$
Thus, the solution to $x^2+2x-9\ge 0$ is:
$x\in \left(-\infty ,-1-\sqrt{10}\right)\cup \left[-1+\sqrt{10},\infty \right]$

Step 5: Applying constraints
Now we intersect this solution set with the initial constraints:
1. $x\ne \pm 2$
2. $x\ne \frac{5}{2}$

Let's approximate the critical values to see how they lie relative to our constraints:
- $-1-\sqrt{10}\approx -1-3.16=-4.16$. Since this is less than $-2$, the value $-2$ is excluded from the right interval but does not affect the interval $\left(-\infty ,-1-\sqrt{10}\right]$.
- $-1+\sqrt{10}\approx -1+3.16=2.16$. This is greater than $2$, meaning $x=2$ is already naturally excluded from the second interval.
- $\frac{5}{2}=2.5$. Since $2.5>2.16$, $x=\frac{5}{2}$ lies inside the interval $\left[-1+\sqrt{10},\infty \right)$ and must be explicitly excluded.

Excluding $x=\frac{5}{2}$ from the interval $\left[-1+\sqrt{10},\infty \right)$ splits it into:
$\left[-1+\sqrt{10},\frac{5}{2}\right)\cup \left(\frac{5}{2},\infty \right)$
Combining this with the first interval, we get the domain of the function as:
$\left(-\infty ,-1-\sqrt{10}\right)\cup \left(-1+\sqrt{10},\frac{5}{2}\right)\cup \left(\frac{5}{2},\infty \right)$