Question

Show that 1/3-√8 - 1/√8-√7 + 1/√7-√6 - 1/√6-√5 + 1/√5-2=5​

Answer 1

ANSWER:

To Show:

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

Proof:

$\text{We are given that,}\\\\:\longrightarrow\dfrac{1}{3-\sqrt8}-\dfrac{1}{\sqrt8-\sqrt7}+\dfrac{1}{\sqrt7-\sqrt6}-\dfrac{1}{\sqrt6-\sqrt5}+\dfrac{1}{\sqrt5-2}=5\\\\\text{Solving LHS,}\\\\:\implies\dfrac{1}{\sqrt9-\sqrt8}-\dfrac{1}{\sqrt8-\sqrt7}+\dfrac{1}{\sqrt7-\sqrt6}-\dfrac{1}{\sqrt6-\sqrt5}+\dfrac{1}{\sqrt5-\sqrt4}\\\\\text{We will now rationalize each of the terms separately.}\\\\1.\::\implies\dfrac{1}{\sqrt9-\sqrt8}=\dfrac{1}{\sqrt9-\sqrt8}\times\dfrac{\sqrt9+\sqrt8}{\sqrt9+\sqrt8}\\\\:\implies\dfrac{\sqrt9+\sqrt8}{(\sqrt9)^2-(\sqrt8)^2}\:\implies\dfrac{\sqrt9+\sqrt8}{9-8}\\\\:\implies\dfrac{\sqrt9+\sqrt8}{1}\:\implies\sqrt9+\sqrt8- - - -(1)$

$2.\::\implies\dfrac{1}{\sqrt8-\sqrt7}=\dfrac{1}{\sqrt8-\sqrt7}\times\dfrac{\sqrt8+\sqrt7}{\sqrt8+\sqrt7}\\\\:\implies\dfrac{\sqrt8+\sqrt7}{(\sqrt8)^2-(\sqrt7)^2}\:\implies\dfrac{\sqrt8+\sqrt7}{8-7}\\\\:\implies\dfrac{\sqrt8+\sqrt7}{1}\:\implies\sqrt8+\sqrt7- - - -(2)$

$3.\::\implies\dfrac{1}{\sqrt7-\sqrt6}=\dfrac{1}{\sqrt7-\sqrt6}\times\dfrac{\sqrt7+\sqrt6}{\sqrt7+\sqrt6}\\\\:\implies\dfrac{\sqrt7+\sqrt6}{(\sqrt7)^2-(\sqrt6)^2}\:\implies\dfrac{\sqrt7+\sqrt6}{7-6}\\\\:\implies\dfrac{\sqrt7+\sqrt6}{1}\:\implies\sqrt7+\sqrt6- - - -(3)$

$4.\::\implies\dfrac{1}{\sqrt6-\sqrt5}=\dfrac{1}{\sqrt6-\sqrt5}\times\dfrac{\sqrt6+\sqrt5}{\sqrt6+\sqrt5}\\\\:\implies\dfrac{\sqrt6+\sqrt5}{(\sqrt6)^2-(\sqrt5)^2}\:\implies\dfrac{\sqrt6+\sqrt5}{6-5}\\\\:\implies\dfrac{\sqrt6+\sqrt5}{1}\:\implies\sqrt6+\sqrt5- - - -(4)$

$5.\::\implies\dfrac{1}{\sqrt5-\sqrt4}=\dfrac{1}{\sqrt5-\sqrt4}\times\dfrac{\sqrt5+\sqrt4}{\sqrt5+\sqrt4}\\\\:\implies\dfrac{\sqrt5+\sqrt4}{(\sqrt5)^2-(\sqrt4)^2}\:\implies\dfrac{\sqrt5+\sqrt4}{5-4}\\\\:\implies\dfrac{\sqrt5+\sqrt4}{1}\:\implies\sqrt5+\sqrt4- - - -(5)$

$\text{We have,}\\\\:\implies\dfrac{1}{\sqrt9-\sqrt8}-\dfrac{1}{\sqrt8-\sqrt7}+\dfrac{1}{\sqrt7-\sqrt6}-\dfrac{1}{\sqrt6-\sqrt5}+\dfrac{1}{\sqrt5-\sqrt4}\\\\\text{Putting respective values from (1), (2), (3), (4) \& (5),}\\\\:\implies\left(\sqrt9+\sqrt8\right)-\left(\sqrt8+\sqrt7\right)+\left(\sqrt7+\sqrt6\right)-\left(\sqrt6+\sqrt5\right)+\left(\sqrt5+\sqrt4\right)\\\\:\implies\sqrt9+\sqrt8-\sqrt8-\sqrt7+\sqrt7+\sqrt6-\sqrt6-\sqrt5+\sqrt5+\sqrt4\\\\:\implies\sqrt9+\sqrt8\!\!\!\!\!\bigg{/}-\sqrt8\!\!\!\!\!\bigg{/}-\sqrt7\!\!\!\!\!\bigg{/}+\sqrt7\!\!\!\!\!\bigg{/}+\sqrt6\!\!\!\!\!\bigg{/}-\sqrt6\!\!\!\!\!\bigg{/}-\sqrt5\!\!\!\!\!\bigg{/}+\sqrt5\!\!\!\!\!\bigg{/}+\sqrt4$

$:\implies\sqrt9+\sqrt4\\\\:\implies3+2\\\\\bf{:\implies 5=RHS}\\\\\text{\bf{Hence Proved!!!}}$

Formula Used:

• (a + b)(a - b) = (a² - b²)