The value of 3+frac{1}{sqrt{3}}+frac{1}{3+sqr | vedclass

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(B) Given expression: $3+\frac{1}{\sqrt{3}}+\frac{1}{3+\sqrt{3}}+\frac{1}{\sqrt{3}-3}$We can rewrite the last term as: $\frac{1}{\sqrt{3}-3} = -\frac{

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$3$

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(B) Given expression: $3+\frac{1}{\sqrt{3}}+\frac{1}{3+\sqrt{3}}+\frac{1}{\sqrt{3}-3}$We can rewrite the last term as: $\frac{1}{\sqrt{3}-3} = -\frac{1}{3-\sqrt{3}}$So the expression becomes: $3+\frac{1}{\sqrt{3}}+\left(\frac{1}{3+\sqrt{3}}-\frac{1}{3-\sqrt{3}}\right)$Now,simplify the term in the parentheses by taking the common denominator $(3+\sqrt{3})(3-\sqrt{3}) = 3^2 - (\sqrt{3})^2 = 9-3 = 6$:$\frac{(3-\sqrt{3}) - (3+\sqrt{3})}{6} = \frac{3-\sqrt{3}-3-\sqrt{3}}{6} = \frac{-2\sqrt{3}}{6} = -\frac{\sqrt{3}}{3} = -\frac{1}{\sqrt{3}}$Substituting this back into the original expression:$3 + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{3}} = 3$

The value of $3+\frac{1}{\sqrt{3}}+\frac{1}{3+\sqrt{3}}+\frac{1}{\sqrt{3}-3}$ is

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