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(A) To simplify the expression,we rationalize each term individually:Step $1$: Rationalize $\frac{2+\sqrt{3}}{2-\sqrt{3}}$:$\frac{2+\sqrt{3}}{2-\sqrt{

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$16-\sqrt{3}$

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(A) To simplify the expression,we rationalize each term individually:Step $1$: Rationalize $\frac{2+\sqrt{3}}{2-\sqrt{3}}$:$\frac{2+\sqrt{3}}{2-\sqrt{3}} 	imes \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{(2+\sqrt{3})^2}{4-3} = 4+3+4\sqrt{3} = 7+4\sqrt{3}$Step $2$: Rationalize $\frac{2-\sqrt{3}}{2+\sqrt{3}}$:$\frac{2-\sqrt{3}}{2+\sqrt{3}} 	imes \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{(2-\sqrt{3})^2}{4-3} = 4+3-4\sqrt{3} = 7-4\sqrt{3}$Step $3$: Rationalize $\frac{\sqrt{3}-1}{\sqrt{3}+1}$:$\frac{\sqrt{3}-1}{\sqrt{3}+1} 	imes \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{3+1-2\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$Step $4$: Add the results:$(7+4\sqrt{3}) + (7-4\sqrt{3}) + (2-\sqrt{3})$$= 7 + 7 + 2 + 4\sqrt{3} - 4\sqrt{3} - \sqrt{3}$$= 16 - \sqrt{3}$

$\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$ simplifies to

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