If frac{1}{N} = frac{(sqrt{6} + sqrt{5})}{(sq | vedclass

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(C) Given that $\frac{1}{N} = \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} - \sqrt{5}}.$To find $N,$ we take the reciprocal of both sides: $N = \frac{\sqrt{6}

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$11 - 2\sqrt{30}$

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(C) Given that $\frac{1}{N} = \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} - \sqrt{5}}.$To find $N,$ we take the reciprocal of both sides: $N = \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}.$Now,rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{6} - \sqrt{5})$:$N = \frac{(\sqrt{6} - \sqrt{5})(\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5})(\sqrt{6} - \sqrt{5})}$Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$ for the numerator and $(a + b)(a - b) = a^2 - b^2$ for the denominator:$N = \frac{(\sqrt{6})^2 + (\sqrt{5})^2 - 2(\sqrt{6})(\sqrt{5})}{(\sqrt{6})^2 - (\sqrt{5})^2}$$N = \frac{6 + 5 - 2\sqrt{30}}{6 - 5}$$N = \frac{11 - 2\sqrt{30}}{1}$$N = 11 - 2\sqrt{30}.$

If $\frac{1}{N} = \frac{(\sqrt{6} + \sqrt{5})}{(\sqrt{6} - \sqrt{5})},$ what is the value of $N?$

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