If `N=(sqrt(sqrt(5)+2)+sqrt(sqrt(5)-2))/(sqrt(sqrt(5)+1))-sqrt(3-2sqrt(2)` then `sqrtN` equals____

Options

A. 1
B. `2sqrt(2)-1`
C. `(sqrt(5))/2`
D. `2/(sqrt(sqrt(5)+1))`

Answer

To solve the problem, we need to simplify the expression for \( N \) given by:

\[ N = \frac{\sqrt{\sqrt{5}+2} + \sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}} - \sqrt{3 - 2\sqrt{2}} \]

### Step 1: Simplify the term \( \sqrt{3 - 2\sqrt{2}} \)

We can rewrite \( 3 - 2\sqrt{2} \) as a square:

\[ 3 - 2\sqrt{2} = (\sqrt{2} - 1)^2 \]

Thus, we have:

\[ \sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1 \]

### Step 2: Substitute back into the expression for \( N \)

Now substituting this back into \( N \):

\[ N = \frac{\sqrt{\sqrt{5}+2} + \sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}} - (\sqrt{2} - 1) \]

This simplifies to:

\[ N = \frac{\sqrt{\sqrt{5}+2} + \sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}} - \sqrt{2} + 1 \]

### Step 3: Simplify the numerator \( \sqrt{\sqrt{5}+2} + \sqrt{\sqrt{5}-2} \)

Let’s denote:

\[ a = \sqrt{\sqrt{5}+2}, \quad b = \sqrt{\sqrt{5}-2} \]

Now, we can find \( a^2 + b^2 \) and \( ab \):

\[ a^2 = \sqrt{5} + 2, \quad b^2 = \sqrt{5} - 2 \]

Adding these gives:

\[ a^2 + b^2 = (\sqrt{5} + 2) + (\sqrt{5} - 2) = 2\sqrt{5} \]

Now, for \( ab \):

\[ ab = \sqrt{(\sqrt{5}+2)(\sqrt{5}-2)} = \sqrt{5 - 4} = 1 \]

### Step 4: Calculate \( a + b \)

Using the identity \( (a+b)^2 = a^2 + b^2 + 2ab \):

\[ (a+b)^2 = 2\sqrt{5} + 2 = 2(\sqrt{5} + 1) \]

Thus, we have:

\[ a + b = \sqrt{2(\sqrt{5}+1)} \]

### Step 5: Substitute \( a + b \) back into \( N \)

Now substituting \( a + b \) back into \( N \):

\[ N = \frac{\sqrt{2(\sqrt{5}+1)}}{\sqrt{\sqrt{5}+1}} - \sqrt{2} + 1 \]

This simplifies to:

\[ N = \sqrt{2} - \sqrt{2} + 1 = 1 \]

### Step 6: Find \( \sqrt{N} \)

Finally, we find:

\[ \sqrt{N} = \sqrt{1} = 1 \]

Thus, the value of \( \sqrt{N} \) is:

\[ \boxed{1} \]