Question

9.Prove that 13−√8+1√8−√7+1√7−√6+1√6−√5+1√5−2=5.

Answer 1

QUESTION:

• Prove that: 1/(3 - √8) - 1/(√8 - √7) + 1/(√7 - √6) - 1/(√6 - √5) + 1/(√5 - 2) = 5

ANSWER:

TO PROVE:

• 1/(3 - √8) - 1/(√8 - √7) + 1/(√7 - √6) - 1/(√6 - √5) + 1/(√5 - 2) = 5

PROOF:

We need to prove that,

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

Solving LHS,

$:\implies\sf\dfrac{1}{3-\sqrt8}-\dfrac{1}{\sqrt8-\sqrt7}+\dfrac{1}{\sqrt7-\sqrt6}-\dfrac{1}{\sqrt6-\sqrt5}+\dfrac{1}{\sqrt5-2}$

We know that,

⇒ 3 = √9 and 2 = √4.

So,

$:\implies\sf\dfrac{1}{\sqrt9-\sqrt8}-\dfrac{1}{\sqrt8-\sqrt7}+\dfrac{1}{\sqrt7-\sqrt6}-\dfrac{1}{\sqrt6-\sqrt5}+\dfrac{1}{\sqrt5-\sqrt4}$

Now, we will rationalise each term individually and then place them in the above equation.

Term 1:

$:\implies\sf\dfrac{1}{\sqrt9-\sqrt8}$

Here the rationalizing factor is (√9 + √8). So,

$:\implies\sf\dfrac{1}{\sqrt9-\sqrt8}\times\dfrac{\sqrt9+\sqrt8}{\sqrt9+\sqrt8}$

$:\implies\sf\dfrac{\sqrt9+\sqrt8}{(\sqrt9-\sqrt8)(\sqrt9+\sqrt8)}$

We know that,

⇒ (a + b)(a - b) = a² - b²

So,

$:\implies\sf\dfrac{\sqrt9+\sqrt8}{(\sqrt9)^2-(\sqrt8)^2}$

$:\implies\sf\dfrac{\sqrt9+\sqrt8}{9-8}$

So,

$:\implies\sf\dfrac{\sqrt9+\sqrt8}{1}$

Hence,

$:\implies\sf\sqrt9+\sqrt8 - - - -(1)$

Term 2:

$:\implies\sf\dfrac{1}{\sqrt8-\sqrt7}$

Here the rationalizing factor is (√8 + √7). So,

$:\implies\sf\dfrac{1}{\sqrt8-\sqrt7}\times\dfrac{\sqrt8+\sqrt7}{\sqrt8+\sqrt7}$

$:\implies\sf\dfrac{\sqrt8+\sqrt7}{(\sqrt8-\sqrt7)(\sqrt8+\sqrt7)}$

We know that,

⇒ (a + b)(a - b) = a² - b²

So,

$:\implies\sf\dfrac{\sqrt8+\sqrt7}{(\sqrt8)^2-(\sqrt7)^2}$

$:\implies\sf\dfrac{\sqrt8+\sqrt7}{8-7}$

So,

$:\implies\sf\dfrac{\sqrt8+\sqrt7}{1}$

Hence,

$:\implies\sf\sqrt8+\sqrt7 - - - -(2)$

Term 3:

$:\implies\sf\dfrac{1}{\sqrt7-\sqrt6}$

Here the rationalizing factor is (√7 + √6). So,

$:\implies\sf\dfrac{1}{\sqrt7-\sqrt6}\times\dfrac{\sqrt7+\sqrt6}{\sqrt7+\sqrt6}$

$:\implies\sf\dfrac{\sqrt7+\sqrt6}{(\sqrt7-\sqrt6)(\sqrt7+\sqrt6)}$

We know that,

⇒ (a + b)(a - b) = a² - b²

So,

$:\implies\sf\dfrac{\sqrt7+\sqrt6}{(\sqrt7)^2-(\sqrt6)^2}$

$:\implies\sf\dfrac{\sqrt7+\sqrt6}{7-6}$

So,

$:\implies\sf\dfrac{\sqrt7+\sqrt6}{1}$

Hence,

$:\implies\sf\sqrt7+\sqrt6 - - - -(3)$

Term 4:

$:\implies\sf\dfrac{1}{\sqrt6-\sqrt5}$

Here the rationalizing factor is (√6 + √5). So,

$:\implies\sf\dfrac{1}{\sqrt6-\sqrt5}\times\dfrac{\sqrt6+\sqrt5}{\sqrt6+\sqrt5}$

$:\implies\sf\dfrac{\sqrt6+\sqrt5}{(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)}$

We know that,

⇒ (a + b)(a - b) = a² - b²

So,

$:\implies\sf\dfrac{\sqrt6+\sqrt5}{(\sqrt6)^2-(\sqrt5)^2}$

$:\implies\sf\dfrac{\sqrt6+\sqrt5}{6-5}$

So,

$:\implies\sf\dfrac{\sqrt6+\sqrt5}{1}$

Hence,

$:\implies\sf\sqrt6+\sqrt5 - - - -(4)$

Term 5:

$:\implies\sf\dfrac{1}{\sqrt5-\sqrt4}$

Here the rationalizing factor is (√5 + √4). So,

$:\implies\sf\dfrac{1}{\sqrt5-\sqrt4}\times\dfrac{\sqrt5+\sqrt4}{\sqrt5+\sqrt4}$

$:\implies\sf\dfrac{\sqrt5+\sqrt4}{(\sqrt5-\sqrt4)(\sqrt5+\sqrt4)}$

We know that,

⇒ (a + b)(a - b) = a² - b²

So,

$:\implies\sf\dfrac{\sqrt5+\sqrt4}{(\sqrt5)^2-(\sqrt4)^2}$

$:\implies\sf\dfrac{\sqrt5+\sqrt4}{5-4}$

So,

$:\implies\sf\dfrac{\sqrt5+\sqrt4}{1}$

Hence,

$:\implies\sf\sqrt5+\sqrt4 - - - -(5)$

We have,

$:\implies\sf\dfrac{1}{\sqrt9-\sqrt8}-\dfrac{1}{\sqrt8-\sqrt7}+\dfrac{1}{\sqrt7-\sqrt6}-\dfrac{1}{\sqrt6-\sqrt5}+\dfrac{1}{\sqrt5-\sqrt4}$

Substituting values from (1), (2), (3), (4) & (5) into above equation,

$:\implies\sf(\sqrt9+\sqrt8)-(\sqrt8+\sqrt7)+(\sqrt7+\sqrt6)-(\sqrt6+\sqrt5)+(\sqrt5+\sqrt4)$

Opening brackets,

$:\implies\sf\sqrt9+\sqrt8\!\!\!\!/\:-\sqrt8\!\!\!\!/\:-\sqrt7\!\!\!\!/\:+\sqrt7\!\!\!\!/\:+\sqrt6\!\!\!\!/\:-\sqrt6\!\!\!\!/\:-\sqrt5\!\!\!\!/\:+\sqrt5\!\!\!\!/\:+\sqrt4$

So,

$:\implies\sf\sqrt9+\sqrt4$

As,

⇒ √9 = 3 and √4 = 2.

So,

$:\implies\sf3+2$

$:\implies\bf{5 = RHS}$

HENCE PROVED!!!