If P = frac{(sqrt{7} - sqrt{6})}{(sqrt{7} + s | vedclass

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(D) Given $P = \frac{\sqrt{7} - \sqrt{6}}{\sqrt{7} + \sqrt{6}}.$ To simplify $P,$ we rationalize the denominator: $P = \frac{(\sqrt{7} - \sqrt{6})(\sq

Vedclass · Accepted Answer

$26$

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(D) Given $P = \frac{\sqrt{7} - \sqrt{6}}{\sqrt{7} + \sqrt{6}}.$ To simplify $P,$ we rationalize the denominator: $P = \frac{(\sqrt{7} - \sqrt{6})(\sqrt{7} - \sqrt{6})}{(\sqrt{7} + \sqrt{6})(\sqrt{7} - \sqrt{6})} = \frac{(\sqrt{7} - \sqrt{6})^2}{7 - 6} = 7 + 6 - 2\sqrt{42} = 13 - 2\sqrt{42}.$ Now,find $\frac{1}{P}$: $\frac{1}{P} = \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} - \sqrt{6}} = \frac{(\sqrt{7} + \sqrt{6})(\sqrt{7} + \sqrt{6})}{(\sqrt{7} - \sqrt{6})(\sqrt{7} + \sqrt{6})} = \frac{(\sqrt{7} + \sqrt{6})^2}{7 - 6} = 7 + 6 + 2\sqrt{42} = 13 + 2\sqrt{42}.$ Finally,calculate $P + \frac{1}{P}$: $P + \frac{1}{P} = (13 - 2\sqrt{42}) + (13 + 2\sqrt{42}) = 13 + 13 = 26.$

If $P = \frac{(\sqrt{7} - \sqrt{6})}{(\sqrt{7} + \sqrt{6})},$ then what is the value of $\left(P + \frac{1}{P}\right)?$

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