The value of left{frac{3 sqrt{2}}{(sqrt{3}+sq | vedclass

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(B) To solve the expression,we rationalize each term individually:$1$. $\frac{3 \sqrt{2}}{\sqrt{6}+\sqrt{3}} = \frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{6-

Vedclass · Accepted Answer

$0$

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(B) To solve the expression,we rationalize each term individually:$1$. $\frac{3 \sqrt{2}}{\sqrt{6}+\sqrt{3}} = \frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3} = \frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{3} = \sqrt{2}(\sqrt{6}-\sqrt{3}) = \sqrt{12}-\sqrt{6} = 2 \sqrt{3}-\sqrt{6}$$2$. $\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}} = \frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2} = \frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{4} = \sqrt{3}(\sqrt{6}-\sqrt{2}) = \sqrt{18}-\sqrt{6} = 3 \sqrt{2}-\sqrt{6}$$3$. $\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}} = \frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2} = \sqrt{6}(\sqrt{3}-\sqrt{2}) = \sqrt{18}-\sqrt{12} = 3 \sqrt{2}-2 \sqrt{3}$Substituting these back into the original expression:$(2 \sqrt{3}-\sqrt{6}) - (3 \sqrt{2}-\sqrt{6}) + (3 \sqrt{2}-2 \sqrt{3})$$= 2 \sqrt{3} - \sqrt{6} - 3 \sqrt{2} + \sqrt{6} + 3 \sqrt{2} - 2 \sqrt{3}$$= (2 \sqrt{3} - 2 \sqrt{3}) + (-\sqrt{6} + \sqrt{6}) + (-3 \sqrt{2} + 3 \sqrt{2})$$= 0 + 0 + 0 = 0$

The value of $\left\{\frac{3 \sqrt{2}}{(\sqrt{3}+\sqrt{6})}-\frac{4 \sqrt{3}}{(\sqrt{6}+\sqrt{2})}+\frac{\sqrt{6}}{(\sqrt{2}+\sqrt{3})}\right\}$ is:

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