The simplified value of the following express | vedclass

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(A) To simplify the expression,we simplify each term individually by converting the nested radicals into the form $\sqrt{(\sqrt{a} \pm \sqrt{b})^2} =

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(A) To simplify the expression,we simplify each term individually by converting the nested radicals into the form $\sqrt{(\sqrt{a} \pm \sqrt{b})^2} = \sqrt{a} \pm \sqrt{b}$.$1$. For the first term: $\frac{1}{\sqrt{11-2 \sqrt{30}}}$.We need two numbers whose sum is $11$ and product is $30$. These are $6$ and $5$.So,$\sqrt{11-2 \sqrt{30}} = \sqrt{(\sqrt{6}-\sqrt{5})^2} = \sqrt{6}-\sqrt{5}$.Thus,$\frac{1}{\sqrt{6}-\sqrt{5}} = \frac{\sqrt{6}+\sqrt{5}}{6-5} = \sqrt{6}+\sqrt{5}$.$2$. For the second term: $\frac{3}{\sqrt{7-2 \sqrt{10}}}$.We need two numbers whose sum is $7$ and product is $10$. These are $5$ and $2$.So,$\sqrt{7-2 \sqrt{10}} = \sqrt{(\sqrt{5}-\sqrt{2})^2} = \sqrt{5}-\sqrt{2}$.Thus,$\frac{3}{\sqrt{5}-\sqrt{2}} = \frac{3(\sqrt{5}+\sqrt{2})}{5-2} = \frac{3(\sqrt{5}+\sqrt{2})}{3} = \sqrt{5}+\sqrt{2}$.$3$. For the third term: $\frac{4}{\sqrt{8+4 \sqrt{3}}} = \frac{4}{\sqrt{8+2 \sqrt{12}}}$.We need two numbers whose sum is $8$ and product is $12$. These are $6$ and $2$.So,$\sqrt{8+2 \sqrt{12}} = \sqrt{(\sqrt{6}+\sqrt{2})^2} = \sqrt{6}+\sqrt{2}$.Thus,$\frac{4}{\sqrt{6}+\sqrt{2}} = \frac{4(\sqrt{6}-\sqrt{2})}{6-2} = \frac{4(\sqrt{6}-\sqrt{2})}{4} = \sqrt{6}-\sqrt{2}$.Substituting these 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$.

The simplified value of the following expression is: $\frac{1}{\sqrt{11-2 \sqrt{30}}}-\frac{3}{\sqrt{7-2 \sqrt{10}}}-\frac{4}{\sqrt{8+4 \sqrt{3}}}$

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