left[frac{3 sqrt{2}}{sqrt{6}-sqrt{3}}-frac{4 | vedclass

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Question

(C) To solve the expression $\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}-\frac{6}{\sqrt{8}-\sqrt{12}}$,we rationalize ea

Vedclass · Accepted Answer

$5 \sqrt{3}$

Vedclass · Answer

(C) To solve the expression $\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}-\frac{6}{\sqrt{8}-\sqrt{12}}$,we rationalize each term individually.First term: $\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}} 	imes \frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}} = \frac{3 \sqrt{12}+3 \sqrt{6}}{6-3} = \frac{6 \sqrt{3}+3 \sqrt{6}}{3} = 2 \sqrt{3}+\sqrt{6}$.Second term: $\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}} 	imes \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}} = \frac{4 \sqrt{18}+4 \sqrt{6}}{6-2} = \frac{12 \sqrt{2}+4 \sqrt{6}}{4} = 3 \sqrt{2}+\sqrt{6}$.Third term: $\frac{6}{\sqrt{8}-\sqrt{12}} 	imes \frac{\sqrt{8}+\sqrt{12}}{\sqrt{8}+\sqrt{12}} = \frac{6(\sqrt{8}+\sqrt{12})}{8-12} = \frac{6(2 \sqrt{2}+2 \sqrt{3})}{-4} = -\frac{3(2 \sqrt{2}+2 \sqrt{3})}{2} = -3 \sqrt{2}-3 \sqrt{3}$.Combining these: $(2 \sqrt{3}+\sqrt{6}) - (3 \sqrt{2}+\sqrt{6}) - (-3 \sqrt{2}-3 \sqrt{3})$.$= 2 \sqrt{3}+\sqrt{6}-3 \sqrt{2}-\sqrt{6}+3 \sqrt{2}+3 \sqrt{3}$.$= 2 \sqrt{3}+3 \sqrt{3} = 5 \sqrt{3}$.

$\left[\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}-\frac{6}{\sqrt{8}-\sqrt{12}}\right] = ?$

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