If `N=(sqrt""9+sqrt""7)div(sqrt""9-sqrt""7)`, then what is the simplified value of 1/N?

Options

A. `8-sqrt""63`
B. `8+sqrt""63`
C. `4+sqrt""17`
D. `4-sqrt""17`

Answer

To solve the problem, we need to find the simplified value of \( \frac{1}{N} \) where \( N = \frac{\sqrt{9} + \sqrt{7}}{\sqrt{9} - \sqrt{7}} \).

### Step-by-Step Solution:

1. **Identify the expression for \( N \)**:
\[ N = \frac{\sqrt{9} + \sqrt{7}}{\sqrt{9} - \sqrt{7}} \]

2. **Find \( \frac{1}{N} \)**:
Taking the reciprocal of \( N \):
\[ \frac{1}{N} = \frac{\sqrt{9} - \sqrt{7}}{\sqrt{9} + \sqrt{7}} \]

3. **Rationalize the denominator**:
To simplify \( \frac{1}{N} \), we multiply the numerator and denominator by \( \sqrt{9} - \sqrt{7} \):
\[ \frac{1}{N} = \frac{(\sqrt{9} - \sqrt{7})(\sqrt{9} - \sqrt{7})}{(\sqrt{9} + \sqrt{7})(\sqrt{9} - \sqrt{7})} \]

4. **Simplify the numerator**:
The numerator becomes:
\[ (\sqrt{9} - \sqrt{7})^2 = \sqrt{9}^2 - 2\sqrt{9}\sqrt{7} + \sqrt{7}^2 = 9 - 2\sqrt{63} + 7 = 16 - 2\sqrt{63} \]

5. **Simplify the denominator**:
The denominator can be simplified using the difference of squares:
\[ (\sqrt{9})^2 - (\sqrt{7})^2 = 9 - 7 = 2 \]

6. **Combine the results**:
Thus, we have:
\[ \frac{1}{N} = \frac{16 - 2\sqrt{63}}{2} \]

7. **Final simplification**:
Dividing both terms in the numerator by 2:
\[ \frac{1}{N} = 8 - \sqrt{63} \]

### Conclusion:
The simplified value of \( \frac{1}{N} \) is:
\[ \frac{1}{N} = 8 - \sqrt{63} \]