If `N=(sqrt(sqrt(5)+2)+sqrt(sqrt(5)-2))/(sqrt(sqrt(5)+1))-sqrt(3-2sqrt(2))` , then `N+2` equals

Answer

To solve the expression \( N = \frac{\sqrt{\sqrt{5}+2} + \sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}} - \sqrt{3 - 2\sqrt{2}} \) and find \( N + 2 \), we will follow these steps:

### Step 1: Simplify the numerator
Let:
\[ a = \sqrt{\sqrt{5}+2} \quad \text{and} \quad b = \sqrt{\sqrt{5}-2} \]
Then the numerator becomes \( a + b \).

### Step 2: Square the numerator
To simplify \( a + b \), we will square it:
\[ (a + b)^2 = a^2 + b^2 + 2ab \]
Calculating \( a^2 \) and \( b^2 \):
\[ a^2 = \sqrt{5} + 2 \quad \text{and} \quad b^2 = \sqrt{5} - 2 \]
Thus,
\[ a^2 + b^2 = (\sqrt{5} + 2) + (\sqrt{5} - 2) = 2\sqrt{5} \]
Now, we need to find \( ab \):
\[ ab = \sqrt{(\sqrt{5}+2)(\sqrt{5}-2)} = \sqrt{5 - 4} = 1 \]
So,
\[ 2ab = 2 \cdot 1 = 2 \]
Therefore,
\[ (a + b)^2 = 2\sqrt{5} + 2 \]

### Step 3: Rewrite the numerator
Now we can express the numerator:
\[ a + b = \sqrt{2\sqrt{5} + 2} \]

### Step 4: Simplify the denominator
The denominator is:
\[ \sqrt{\sqrt{5}+1} \]

### Step 5: Substitute back into \( N \)
Now substituting back into \( N \):
\[ N = \frac{\sqrt{2\sqrt{5} + 2}}{\sqrt{\sqrt{5}+1}} - \sqrt{3 - 2\sqrt{2}} \]

### Step 6: Simplify \( \sqrt{3 - 2\sqrt{2}} \)
Notice that:
\[ 3 - 2\sqrt{2} = (\sqrt{2} - 1)^2 \]
Thus,
\[ \sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1 \]

### Step 7: Substitute and simplify \( N \)
Now substituting this back into \( N \):
\[ N = \frac{\sqrt{2\sqrt{5} + 2}}{\sqrt{\sqrt{5}+1}} - (\sqrt{2} - 1) \]
This can be simplified further, but for \( N + 2 \):
\[ N + 2 = \frac{\sqrt{2\sqrt{5} + 2}}{\sqrt{\sqrt{5}+1}} - \sqrt{2} + 1 + 2 \]
\[ N + 2 = \frac{\sqrt{2\sqrt{5} + 2}}{\sqrt{\sqrt{5}+1}} - \sqrt{2} + 3 \]

### Step 8: Final simplification
After careful simplification, we find that:
\[ N + 2 = 3 \]

Thus, the final answer is:
\[ \boxed{3} \]