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

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Question

(B) Rationalizing each term separately:$1$. $\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}} = \frac{3 \sqrt{2}(\sqrt{6}+\sqrt{3})}{6-3} = \frac{3(\sqrt{12}+\sqr

Vedclass · Accepted Answer

$-\sqrt{3}$

Vedclass · Answer

(B) Rationalizing each term separately:$1$. $\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}} = \frac{3 \sqrt{2}(\sqrt{6}+\sqrt{3})}{6-3} = \frac{3(\sqrt{12}+\sqrt{6})}{3} = 2 \sqrt{3}+\sqrt{6}$$2$. $\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}} = \frac{4 \sqrt{3}(\sqrt{6}+\sqrt{2})}{6-2} = \frac{4(\sqrt{18}+\sqrt{6})}{4} = 3 \sqrt{2}+\sqrt{6}$$3$. $\frac{6}{\sqrt{8}+\sqrt{12}} = \frac{6(\sqrt{12}-\sqrt{8})}{12-8} = \frac{6(2 \sqrt{3}-2 \sqrt{2})}{4} = \frac{3(2 \sqrt{3}-2 \sqrt{2})}{2} = 3 \sqrt{3}-3 \sqrt{2}$Substituting these values into the original expression:$(2 \sqrt{3}+\sqrt{6}) - (3 \sqrt{2}+\sqrt{6}) - (3 \sqrt{3}-3 \sqrt{2})$$= 2 \sqrt{3} + \sqrt{6} - 3 \sqrt{2} - \sqrt{6} - 3 \sqrt{3} + 3 \sqrt{2}$$= (2 \sqrt{3} - 3 \sqrt{3}) + (\sqrt{6} - \sqrt{6}) + (-3 \sqrt{2} + 3 \sqrt{2})$$= -\sqrt{3}$

$\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}-\frac{6}{\sqrt{8}+\sqrt{12}}=?$

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