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Ch - Number Systems
Class 9th
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Answer 1

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Answer 2

Given Question: Prove that:

$\sf{\dfrac{1}{3-\sqrt8}-\dfrac{1}{\sqrt8-\sqrt7}+\dfrac{1}{\sqrt7-\sqrt6}-\dfrac{1}{\sqrt6-\sqrt5}+\dfrac{1}{\sqrt5-2}=5}$

Solution:

All the fractions are having irrational numbers as denominators. Before we move on, we need to rationalize the denominators.

$\sf{\bigg(\dfrac{1}{3-\sqrt8}\times\dfrac{3+\sqrt8}{3+\sqrt8}\bigg)$$-\bigg(\dfrac{1}{\sqrt8-\sqrt7}\times\dfrac{\sqrt8+\sqrt7}{\sqrt8+\sqrt7}\bigg)$$+\bigg(\dfrac{1}{\sqrt7-\sqrt6}\times\dfrac{\sqrt7+\sqrt6}{\sqrt7+\sqrt6}\bigg)$$-\bigg(\dfrac{1}{\sqrt6\sqrt5}\times\dfrac{\sqrt6+\sqrt5}{\sqrt6+\sqrt5}\bigg)+\bigg(\dfrac{1}{\sqrt5-2}\times\dfrac{\sqrt5+2}{\sqrt5+2}\bigg)=5}$

Now solving the LHS of the statement:

$\sf{=\dfrac{3+\sqrt8}{9-8}-\dfrac{\sqrt8+\sqrt7}{8-7}+\dfrac{\sqrt7+\sqrt6}{7-6}-\dfrac{\sqrt6+\sqrt5}{6-5}+\dfrac{\sqrt5+2}{5-4}}$

$\sf{=3+\sqrt8-\sqrt8-\sqrt7+\sqrt7+\sqrt6-\sqrt6-\sqrt5+\sqrt5+2}$

Now, cancelling the terms with opposite signs, we are left with:

$\sf{=3+2}$

$\sf{=5}$

The RHS of the statement is:

= 5

LHS = RHS

Hence proved!

Rationalizing the denominator:

If we are given fractions with irrational numbers as denominators, we have to rationalize their denominators by multiplying both the numerator and denominator with the term of the denominator with the opposite sign.