A) \[\sqrt{6}\]
B) \[\sqrt{3}\]
C) \[\sqrt{2}\]
D) \[0\]
Correct Answer: D
Solution :
| \[\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}=\frac{3\sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}\] (Rationalising the denominator] |
| \[=\frac{3\sqrt{12}-3\sqrt{6}}{6-3}=\frac{3(2\sqrt{3}-\sqrt{6})}{3}=2\sqrt{3}-\sqrt{6}\] |
| Similarly, \[\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}=\frac{4\sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}\] |
| \[=\frac{4\sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}=\sqrt{3}(\sqrt{6}-\sqrt{2})\] |
| \[=\sqrt{18}-\sqrt{6}=3\sqrt{2}-\sqrt{6}\] |
| \[\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\] |
| \[=\frac{\sqrt{18}-\sqrt{12}}{3-2}=3\sqrt{2}-2\sqrt{3}\] |
| \[\therefore \] Expression |
| \[=2\sqrt{3}-\sqrt{6}-3\sqrt{2}+\sqrt{6}+3\sqrt{2}-2\sqrt{3}\] |
| \[=0\] |
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