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

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Question

(A) To rationalize the denominator,we write the expression as $\frac{4}{(1 + \sqrt{2}) - \sqrt{3}}$.Multiply the numerator and denominator by $(1 + \s

Vedclass · Accepted Answer

$2 + \sqrt 2 + \sqrt 6 $

Vedclass · Answer

(A) To rationalize the denominator,we write the expression as $\frac{4}{(1 + \sqrt{2}) - \sqrt{3}}$.Multiply the numerator and denominator by $(1 + \sqrt{2}) + \sqrt{3}$:$\frac{4((1 + \sqrt{2}) + \sqrt{3})}{((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}$.

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

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