Question: Find out the value of $ \frac{1}{\sqrt{9}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}} $ $ -\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{4}} $
Options:
A) 0
B) 5
C) 7
D) 8
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \frac{1}{\sqrt{9}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}} $
$ -\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{4}} $
$ \therefore $ $ \frac{1}{\sqrt{9}-\sqrt{8}}\times \frac{(\sqrt{9}+\sqrt{8})}{(\sqrt{9}+\sqrt{8})}=\frac{\sqrt{9}+\sqrt{8}}{9-8}=\sqrt{9}+\sqrt{8} $
Similarly,
$ \frac{1}{\sqrt{8}-\sqrt{7}}=\sqrt{8}+\sqrt{7}; $ $ \frac{1}{\sqrt{7}-\sqrt{6}}=\sqrt{7}+\sqrt{6} $
$ \frac{1}{\sqrt{6}-\sqrt{5}}=\sqrt{6}+\sqrt{5} $ and $ \frac{1}{\sqrt{5}-\sqrt{4}}=\sqrt{5}+\sqrt{4} $
$ \therefore $ Above given expression can be written as,
$ \sqrt{9}+\sqrt{8}-(\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6}) $
$ -(\sqrt{6}+\sqrt{5})+(\sqrt{5}+\sqrt{4}) $
$ \Rightarrow $ $ \sqrt{9}+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+\sqrt{4} $
$ \Rightarrow $ $ \sqrt{9}+\sqrt{4}=3+2=5 $
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