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

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(D) To simplify the expression,we rationalize the denominator for each term:Term $1$: $\frac{3\sqrt{2}}{\sqrt{6} + \sqrt{3}} 	imes \frac{\sqrt{6} - \

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(D) To simplify the expression,we rationalize the denominator for each term:Term $1$: $\frac{3\sqrt{2}}{\sqrt{6} + \sqrt{3}} 	imes \frac{\sqrt{6} - \sqrt{3}}{\sqrt{6} - \sqrt{3}} = \frac{3\sqrt{2}(\sqrt{6} - \sqrt{3})}{6 - 3} = \frac{3\sqrt{2}(\sqrt{6} - \sqrt{3})}{3} = \sqrt{2}(\sqrt{6} - \sqrt{3}) = \sqrt{12} - \sqrt{6} = 2\sqrt{3} - \sqrt{6}$Term $2$: $\frac{4\sqrt{3}}{\sqrt{6} + \sqrt{2}} 	imes \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \frac{4\sqrt{3}(\sqrt{6} - \sqrt{2})}{6 - 2} = \frac{4\sqrt{3}(\sqrt{6} - \sqrt{2})}{4} = \sqrt{3}(\sqrt{6} - \sqrt{2}) = \sqrt{18} - \sqrt{6} = 3\sqrt{2} - \sqrt{6}$Term $3$: $\frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}} 	imes \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{6}(\sqrt{3} - \sqrt{2})}{3 - 2} = \sqrt{6}(\sqrt{3} - \sqrt{2}) = \sqrt{18} - \sqrt{12} = 3\sqrt{2} - 2\sqrt{3}$Combining the terms:$(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} = 0$

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

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