frac{12}{3 + sqrt{5} - 2sqrt{2}} = | vedclass

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Question

(C) Rationalize the denominator by grouping terms: $\frac{12}{(3 - 2\sqrt{2}) + \sqrt{5}} 	imes \frac{(3 - 2\sqrt{2}) - \sqrt{5}}{(3 - 2\sqrt{2}) - \

Vedclass · Accepted Answer

$1 + \sqrt{5} + \sqrt{10} - \sqrt{2}$

Vedclass · Answer

(C) Rationalize the denominator by grouping terms: $\frac{12}{(3 - 2\sqrt{2}) + \sqrt{5}} 	imes \frac{(3 - 2\sqrt{2}) - \sqrt{5}}{(3 - 2\sqrt{2}) - \sqrt{5}}$$= \frac{12(3 - 2\sqrt{2} - \sqrt{5})}{(3 - 2\sqrt{2})^2 - 5} = \frac{12(3 - 2\sqrt{2} - \sqrt{5})}{9 + 8 - 12\sqrt{2} - 5} = \frac{12(3 - 2\sqrt{2} - \sqrt{5})}{12 - 12\sqrt{2}}$$= \frac{3 - 2\sqrt{2} - \sqrt{5}}{1 - \sqrt{2}} = \frac{-(3 - 2\sqrt{2} - \sqrt{5})}{\sqrt{2} - 1} 	imes \frac{\sqrt{2} + 1}{\sqrt{2} + 1}$$= \frac{(-3 + 2\sqrt{2} + \sqrt{5})(\sqrt{2} + 1)}{2 - 1} = -3\sqrt{2} - 3 + 4 + 2\sqrt{2} + \sqrt{10} + \sqrt{5}$$= 1 + \sqrt{5} + \sqrt{10} - \sqrt{2}$.

$\frac{12}{3 + \sqrt{5} - 2\sqrt{2}} = $

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