Evaluate: frac{4}{1 + sqrt{2} - sqrt{3}} | vedclass

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Question

(A) To rationalize the denominator,we multiply the numerator and denominator by $(1 + \sqrt{2}) + \sqrt{3}$:$\frac{4}{(1 + \sqrt{2}) - \sqrt{3}} 	ime

Vedclass · Accepted Answer

$2 + \sqrt{2} + \sqrt{6}$

Vedclass · Answer

(A) To rationalize the denominator,we multiply the numerator and denominator by $(1 + \sqrt{2}) + \sqrt{3}$:$\frac{4}{(1 + \sqrt{2}) - \sqrt{3}} 	imes \frac{(1 + \sqrt{2}) + \sqrt{3}}{(1 + \sqrt{2}) + \sqrt{3}}$$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{(1 + \sqrt{2})^2 - (\sqrt{3})^2}$$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{1 + 2 + 2\sqrt{2} - 3}$$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{2\sqrt{2}}$$= \frac{2(1 + \sqrt{2} + \sqrt{3})}{\sqrt{2}}$$= \sqrt{2}(1 + \sqrt{2} + \sqrt{3})$$= \sqrt{2} + 2 + \sqrt{6}$$= 2 + \sqrt{2} + \sqrt{6}$

Evaluate: $\frac{4}{1 + \sqrt{2} - \sqrt{3}}$

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