A) \[\sqrt{7}-\sqrt{5}\]
B) \[\sqrt{5}-\sqrt{3}\]
C) \[\sqrt{9}-\sqrt{7}\]
D) \[\sqrt{11}-\sqrt{9}\]
Correct Answer: B
Solution :
| \[\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{7}+\sqrt{5}}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}\] |
| \[=\frac{\sqrt{7}+\sqrt{5}}{7-5}=\frac{\sqrt{7}+\sqrt{5}}{2};\frac{1}{\sqrt{5}-\sqrt{3}}\] |
| \[=\frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{\sqrt{5}+\sqrt{3}}{5-3}=\frac{\sqrt{5}+\sqrt{3}}{2}\] |
| Similarly, |
| \[\frac{1}{\sqrt{9}-\sqrt{7}}=\frac{\sqrt{9}+\sqrt{7}}{2}=\frac{1}{\sqrt{11}-\sqrt{9}}\] |
| \[=\frac{\sqrt{11}+\sqrt{9}}{2}\] |
| Clearly, \[\frac{\sqrt{5}+\sqrt{3}}{2}\]is the smallest. |
| \[\therefore \]\[\frac{1}{\sqrt{5}-\sqrt{3}}\]is the smallest. |
| \[\therefore \]\[\sqrt{5}-\sqrt{3}\]is the greatest. |
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