Let 'n' be the number of elements in the domain set of the function `f(x)=abs(ln sqrt(""^(x^(2)+4x)C_(2x^(2)+3)))` and 'Y' be the global maximum value of f(x), then [n+[Y]] is ………. (where `[*]`=greatest integer function).

Answer

To solve the problem, we need to analyze the function \( f(x) = \left| \ln \sqrt{C_{(x^2 + 4x)}^{(2x^2 + 3)}} \right| \) and determine the number of elements in its domain (denoted as \( n \)) and the global maximum value of \( f(x) \) (denoted as \( Y \)). Finally, we will compute \( n + [Y] \), where \( [*] \) denotes the greatest integer function.

### Step-by-Step Solution:

1. **Identify the conditions for the logarithm:**
The logarithm function \( \ln(a) \) is defined only when \( a > 0 \). Therefore, we need:
\[ C_{(x^2 + 4x)}^{(2x^2 + 3)} > 0 \]
This implies that both \( x^2 + 4x \) and \( 2x^2 + 3 \) must be greater than zero.

2. **Analyze \( x^2 + 4x > 0 \):**
Factor the expression:
\[ x(x + 4) > 0 \]
The critical points are \( x = 0 \) and \( x = -4 \). The intervals to test are:
- \( (-\infty, -4) \)
- \( (-4, 0) \)
- \( (0, \infty) \)

Testing these intervals:
- For \( x < -4 \): Choose \( x = -5 \) → \( (-5)(-1) > 0 \) (True)
- For \( -4 < x < 0 \): Choose \( x = -2 \) → \( (-2)(2) < 0 \) (False)
- For \( x > 0 \): Choose \( x = 1 \) → \( (1)(5) > 0 \) (True)

Thus, \( x^2 + 4x > 0 \) for \( x \in (-\infty, -4) \cup (0, \infty) \).

3. **Analyze \( 2x^2 + 3 > 0 \):**
The expression \( 2x^2 + 3 \) is always positive since \( 2x^2 \) is non-negative and adding 3 ensures it is greater than zero for all \( x \).

4. **Combine the conditions:**
The domain of \( f(x) \) is determined by the first condition:
\[ \text{Domain: } (-\infty, -4) \cup (0, \infty) \]

5. **Count the number of elements in the domain:**
Since the domain consists of two intervals, we can say \( n = 2 \).

6. **Find the global maximum value \( Y \):**
To find the maximum value of \( f(x) \), we need to evaluate \( f(x) \) at critical points within the domain. We can check the behavior of \( f(x) \) as \( x \) approaches the boundaries of the intervals:
- As \( x \to 0^+ \), \( f(x) \to \infty \) (since \( \ln(0) \) approaches negative infinity).
- As \( x \to -4^- \), \( f(x) \to \infty \) (similar reasoning).

Therefore, the global maximum value \( Y \) is infinite, but we need to consider the greatest integer function. Since \( Y \) approaches infinity, we can denote \( [Y] = \infty \).

7. **Final computation:**
Now we compute \( n + [Y] \):
\[ n + [Y] = 2 + \infty = \infty \]

### Conclusion:
The final answer is:
\[ \text{The value of } n + [Y] \text{ is } \infty. \]