The number `N=(1+2log_(3)2)/((1+log_(3)2)^(2))+(log_(6)2)^(2)` when simplified reduces to:

Options

A. A prime number
B. an irrational number
C. a real which is less than `log_(3)pi`
D. a real which is greater than `log_(7)6`

Explanation

To simplify the expression \( N = \frac{1 + 2 \log_3 2}{(1 + \log_3 2)^2} + (\log_6 2)^2 \), we will follow these steps:

### Step 1: Rewrite \( \log_6 2 \)
Using the change of base formula, we can express \( \log_6 2 \) as:
\[ \log_6 2 = \frac{\log_2 2}{\log_2 6} = \frac{1}{\log_2 6} \]
Now, we can express \( \log_2 6 \) as:
\[ \log_2 6 = \log_2 (2 \cdot 3) = \log_2 2 + \log_2 3 = 1 + \log_2 3 \]
Thus,
\[ \log_6 2 = \frac{1}{1 + \log_2 3} \]

### Step 2: Substitute \( \log_3 2 \)
Next, we can express \( \log_3 2 \) in terms of \( \log_2 3 \):
\[ \log_3 2 = \frac{1}{\log_2 3} \]
Let \( t = \log_2 3 \). Then, we have:
\[ \log_3 2 = \frac{1}{t} \]

### Step 3: Substitute into \( N \)
Now substituting \( \log_3 2 \) and \( \log_6 2 \) into \( N \):
\[ N = \frac{1 + 2 \cdot \frac{1}{t}}{(1 + \frac{1}{t})^2} + \left(\frac{1}{1+t}\right)^2 \]
This simplifies to:
\[ N = \frac{1 + \frac{2}{t}}{\left(1 + \frac{1}{t}\right)^2} + \frac{1}{(1+t)^2} \]

### Step 4: Simplify the first term
The denominator can be simplified:
\[ \left(1 + \frac{1}{t}\right)^2 = \left(\frac{t+1}{t}\right)^2 = \frac{(t+1)^2}{t^2} \]
Thus, the first term becomes:
\[ \frac{1 + \frac{2}{t}}{\frac{(t+1)^2}{t^2}} = \frac{(1 + \frac{2}{t}) \cdot t^2}{(t+1)^2} = \frac{t^2 + 2t}{(t+1)^2} \]

### Step 5: Combine the terms
Now we can write \( N \) as:
\[ N = \frac{t^2 + 2t}{(t+1)^2} + \frac{1}{(1+t)^2} \]
Finding a common denominator:
\[ N = \frac{t^2 + 2t + 1}{(t+1)^2} \]
This simplifies to:
\[ N = \frac{(t+1)^2}{(t+1)^2} = 1 \]

### Final Result
Thus, the simplified value of \( N \) is:
\[ \boxed{1} \]