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Examveda

$$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - $$  $$\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + $$  $$\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$  is equal to = ?

A. 3

B. 2

C. 0

D. $$\sqrt 3 $$

Answer: Option C

Solution(By Examveda Team)

$$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + \frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$
$$ = \left( {\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} \times \frac{{\sqrt 6 - \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}{\text{ }}} \right) - $$      $$\left( {\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}{\text{ }}} \right) + $$     $$\left( {\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}{\text{ }} \times \frac{{\sqrt 6 - 2}}{{\sqrt 6 - 2}}{\text{ }}} \right)$$
$$ = \frac{{3\sqrt 2 \left( {\sqrt 6 - \sqrt 3 } \right)}}{3} - $$    $$\frac{{2\sqrt 6 \left( {\sqrt 3 - 1} \right)}}{2} + $$   $$\frac{{2\sqrt 3 \left( {\sqrt 6 - 2} \right)}}{2}$$
$$\eqalign{ & = \sqrt {12} - \sqrt 6 - \sqrt {18} + \sqrt 6 + \sqrt {18} - 2\sqrt 3 \cr & = \sqrt {12} - 2\sqrt 3 \cr & = 2\sqrt 3 - 2\sqrt 3 \cr & = 0 \cr} $$

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