If N = frac{sqrt{7} - sqrt{3}}{sqrt{7} + sqrt | vedclass

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(B) Given $N = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}}$.Rationalizing the denominator,we get:$N = \frac{(\sqrt{7} - \sqrt{3})(\sqrt{7} - \sqrt

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$5$

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(B) Given $N = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}}$.Rationalizing the denominator,we get:$N = \frac{(\sqrt{7} - \sqrt{3})(\sqrt{7} - \sqrt{3})}{(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3})} = \frac{7 + 3 - 2\sqrt{21}}{7 - 3} = \frac{10 - 2\sqrt{21}}{4} = \frac{5 - \sqrt{21}}{2}$.Now,$\frac{1}{N} = \frac{2}{5 - \sqrt{21}}$.Rationalizing $\frac{1}{N}$,we get:$\frac{1}{N} = \frac{2(5 + \sqrt{21})}{(5 - \sqrt{21})(5 + \sqrt{21})} = \frac{2(5 + \sqrt{21})}{25 - 21} = \frac{2(5 + \sqrt{21})}{4} = \frac{5 + \sqrt{21}}{2}$.Therefore,$N + \frac{1}{N} = \frac{5 - \sqrt{21}}{2} + \frac{5 + \sqrt{21}}{2} = \frac{5 - \sqrt{21} + 5 + \sqrt{21}}{2} = \frac{10}{2} = 5$.

If $N = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}}$,then what is the value of $N + \frac{1}{N}$?

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