A) \[\sqrt{6}\]
B) \[\sqrt{3}\]
C) \[\sqrt{2}\]
D) 0
Correct Answer: D
Solution :
| Firstly, we rationalise each term by its respective conjugate. |
| 1st term\[=\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}=\frac{3\sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}=\frac{3\sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3}\] |
| \[=\sqrt{2}(\sqrt{6}-\sqrt{3})=\sqrt{12}-\sqrt{6}=2\sqrt{3}-\sqrt{6}\] |
| 2nd term \[=\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}=\frac{4\sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}\] |
| \[=\frac{4\sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}=\sqrt{6\times 3}-\sqrt{2\times 3}=3\sqrt{2}-\sqrt{6}\]3rd term \[=\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\] |
| \[=\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2}=\sqrt{6\times 3}-\sqrt{6\times 3}=3\sqrt{2}-2\sqrt{3}\]\[\therefore \]The given expression |
| \[=(2\sqrt{3}-\sqrt{6})-(3\sqrt{2}-\sqrt{6})+(3\sqrt{2}-2\sqrt{3})\] |
| \[=2\sqrt{3}-\sqrt{6}-3\sqrt{2}+\sqrt{6}+3\sqrt{2}-2\sqrt{3}=0\] |
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