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The value of (19−8−18−7+17−6−16−5+15−4)(\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}}) is

Solution

✅ Correct Option: 4

Let's look at this sum: 19−8−18−7+17−6−16−5+15−4\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}}

Let's rationalize each denominator:

19−8=9+8(9−8)(9+8)=9+89−8=3+221=3+22\frac{1}{\sqrt{9}-\sqrt{8}} = \frac{\sqrt{9}+\sqrt{8}}{(\sqrt{9}-\sqrt{8})(\sqrt{9}+\sqrt{8})} = \frac{\sqrt{9}+\sqrt{8}}{9-8} = \frac{3+2\sqrt{2}}{1} = 3+2\sqrt{2}

Similarly,

18−7=−(22+7)\frac{1}{\sqrt{8}-\sqrt{7}} = -(2\sqrt{2}+\sqrt{7})

17−6=7+6\frac{1}{\sqrt{7}-\sqrt{6}} = \sqrt{7}+\sqrt{6}

16−5=−(6+5)\frac{1}{\sqrt{6}-\sqrt{5}} = -(\sqrt{6}+\sqrt{5})

15−4=5+2\frac{1}{\sqrt{5}-\sqrt{4}} = \sqrt{5}+2

Adding all terms:

3+22−(22+7)+(7+6)−(6+5)+(5+2)3+2\sqrt{2}-(2\sqrt{2}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5})+(\sqrt{5}+2)

All terms cancel except 5.

Therefore, the value is 5.

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