Question

plz solve fast NCERT 9th​

Answer 1

Topic: Rationalization

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To prove

$\dfrac{1}{\sqrt{1} +\sqrt{2} }+\dfrac{1}{\sqrt{2} +\sqrt{3} }+\dfrac{1}{\sqrt{3}+\sqrt{4} }+\dfrac{1}{\sqrt{4} +\sqrt{5} }+...+\dfrac{1}{\sqrt{8}+\sqrt{9} }=2$

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Solution

What can we observe?

Firstly, let's consider $\dfrac{1}{1+\sqrt{2} }$. To rationalize it we should 1 and $\sqrt{2}$ in the denominator. This is done by identity $(a+b)(a-b)=a^2-b^2$.

To rationalize: $\dfrac{1}{1+\sqrt{2} }$

Multiplying '1' to the fraction,

$= \dfrac{1}{1+\sqrt{2} }\times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }$

$=\dfrac{1-\sqrt{2} }{(1+\sqrt{2} )(1-\sqrt{2} )}$

$=\dfrac{1-\sqrt{2} }{1-2}$

$=\sqrt{2} -1$

We can Observe

We can observe the signs of the two terms are different. We repeat this process on each fraction so we can eliminate most of the terms.

Then,

$=(\sqrt{2} -1)+(\sqrt{3} -\sqrt{2} )+(\sqrt{4} -\sqrt{3} )+...+(\sqrt{9} -\sqrt{8} )$

$=(-1)+\sqrt{9}$

$=-1+3=\boxed{2}$

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More Information

This problem is likened to a telescope, because most of the terms vanish, leaving the first and the last term. (Or, it is called a telescoping series.)