The value of frac{3 sqrt{2}}{sqrt{3}+sqrt{6}} | vedclass

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(B) To solve the expression $\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$,we rational

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(B) To solve the expression $\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$,we rationalize the denominator for each term.First term: $\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})} = \frac{3 \sqrt{6}-3 \sqrt{12}}{3-6} = \frac{3 \sqrt{6}-6 \sqrt{3}}{-3} = -\sqrt{6}+2 \sqrt{3}$.Second term: $\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})} = \frac{4 \sqrt{18}-4 \sqrt{6}}{6-2} = \frac{12 \sqrt{2}-4 \sqrt{6}}{4} = 3 \sqrt{2}-\sqrt{6}$.Third term: $\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} = \frac{\sqrt{18}-\sqrt{12}}{3-2} = 3 \sqrt{2}-2 \sqrt{3}$.Combining these: $(-\sqrt{6}+2 \sqrt{3}) - (3 \sqrt{2}-\sqrt{6}) + (3 \sqrt{2}-2 \sqrt{3})$.$= -\sqrt{6}+2 \sqrt{3}-3 \sqrt{2}+\sqrt{6}+3 \sqrt{2}-2 \sqrt{3} = 0$.

The value of $\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$ is

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