frac{1}{sqrt{11 - 2sqrt{30}}} - frac{3}{sqrt{ | vedclass

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Question

(A) We simplify each term individually:$1$. $\frac{1}{\sqrt{11 - 2\sqrt{30}}} = \frac{1}{\sqrt{(\sqrt{6} - \sqrt{5})^2}} = \frac{1}{\sqrt{6} - \sqrt{5

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(A) We simplify each term individually:$1$. $\frac{1}{\sqrt{11 - 2\sqrt{30}}} = \frac{1}{\sqrt{(\sqrt{6} - \sqrt{5})^2}} = \frac{1}{\sqrt{6} - \sqrt{5}} = \sqrt{6} + \sqrt{5}$$2$. $\frac{3}{\sqrt{7 - 2\sqrt{10}}} = \frac{3}{\sqrt{(\sqrt{5} - \sqrt{2})^2}} = \frac{3}{\sqrt{5} - \sqrt{2}} = \frac{3(\sqrt{5} + \sqrt{2})}{5 - 2} = \frac{3(\sqrt{5} + \sqrt{2})}{3} = \sqrt{5} + \sqrt{2}$$3$. $\frac{4}{\sqrt{8 + 4\sqrt{3}}} = \frac{4}{\sqrt{4(2 + \sqrt{3})}} = \frac{4}{2\sqrt{2 + \sqrt{3}}} = \frac{2}{\sqrt{2 + \sqrt{3}}} = \frac{2\sqrt{2 - \sqrt{3}}}{\sqrt{4 - 3}} = 2\sqrt{2 - \sqrt{3}} = \sqrt{8 - 4\sqrt{3}} = \sqrt{6} - \sqrt{2}$Substituting these values back into the expression:$(\sqrt{6} + \sqrt{5}) - (\sqrt{5} + \sqrt{2}) - (\sqrt{6} - \sqrt{2})$$= \sqrt{6} + \sqrt{5} - \sqrt{5} - \sqrt{2} - \sqrt{6} + \sqrt{2} = 0$

$\frac{1}{\sqrt{11 - 2\sqrt{30}}} - \frac{3}{\sqrt{7 - 2\sqrt{10}}} - \frac{4}{\sqrt{8 + 4\sqrt{3}}} = $

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