To find the domain of the function
\[
f(x) = \log_{3} \log_{\frac{1}{3}}(x^{2} + 10x + 25) + \frac{1}{[x] + 5}
\]
we need to consider the restrictions imposed by the logarithmic functions and the greatest integer function.
### Step 1: Analyze the inner logarithm
The inner logarithm is
\[
\log_{\frac{1}{3}}(x^{2} + 10x + 25).
\]
For this logarithm to be defined, the argument must be positive:
\[
x^{2} + 10x + 25 > 0.
\]
### Step 2: Factor the quadratic expression
The expression \(x^{2} + 10x + 25\) can be factored as:
\[
(x + 5)^{2}.
\]
Since a square is always non-negative, we have:
\[
(x + 5)^{2} \geq 0.
\]
This expression is equal to zero when \(x + 5 = 0\), or \(x = -5\). Therefore, the expression is positive for all \(x \neq -5\).
### Step 3: Analyze the base of the logarithm
Next, we need to ensure that the logarithm does not evaluate to 1, since \(\log_{\frac{1}{3}}(1) = 0\) and this would make the outer logarithm undefined. Thus, we set:
\[
x^{2} + 10x + 25 \neq 1.
\]
### Step 4: Solve the inequality
Setting the quadratic equal to 1 gives:
\[
x^{2} + 10x + 25 - 1 = 0 \implies x^{2} + 10x + 24 = 0.
\]
We can factor this as:
\[
(x + 6)(x + 4) = 0.
\]
Thus, \(x = -6\) or \(x = -4\). Therefore, \(x\) cannot equal -6 or -4.
### Step 5: Analyze the greatest integer function
The term \(\frac{1}{[x] + 5}\) requires that the denominator cannot be zero:
\[
[x] + 5 \neq 0 \implies [x] \neq -5.
\]
This means \(x\) cannot be in the interval \([-5, -4)\) because for any \(x\) in this interval, \([x] = -5\).
### Step 6: Combine the restrictions
From the analysis, we have the following restrictions:
1. \(x \neq -5\)
2. \(x \neq -6\)
3. \(x \neq -4\)
4. \(x \in (-\infty, -5) \cup (-5, -4) \cup (-4, \infty)\)
### Final Domain
Thus, the domain of the function \(f(x)\) can be expressed as:
\[
(-\infty, -6) \cup (-6, -5) \cup (-5, -4) \cup (-4, \infty).
\]
