Question

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

Answer 1

Answer:

5+2[

$[7y - 5]$

Mark as brainlest

Answer 2

Answer:

Step-by-step explanation:

Given

$\cfrac{1}{3-2\sqrt{2}}-\cfrac{1}{2\sqrt{2}-\sqrt{7}}+\cfrac{1}{\sqrt{7}-\sqrt{6}}-\cfrac{1}{\sqrt{6}-\sqrt{5}}+\cfrac{1}{\sqrt{5}-2} = 5$

L.H.S.

$\cfrac{1}{3-2\sqrt{2}}-\cfrac{1}{2\sqrt{2}-\sqrt{7}}+\cfrac{1}{\sqrt{7}-\sqrt{6}}-\cfrac{1}{\sqrt{6}-\sqrt{5}}+\cfrac{1}{\sqrt{5}-2}$

$=\cfrac{1}{3-\sqrt{8}}-\cfrac{1}{\sqrt{8}-\sqrt{7}}+\cfrac{1}{\sqrt{7}-\sqrt{6}}-\cfrac{1}{\sqrt{6}-\sqrt{5}}+\cfrac{1}{\sqrt{5}-2}$

Rationalising each term,

$=\cfrac{1}{3-\sqrt{8}}\times\cfrac{3+\sqrt{8}}{3+\sqrt{8}}-\cfrac{1}{\sqrt{8}-\sqrt{7}}\times \cfrac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}+\cfrac{1}{\sqrt{7}-\sqrt{6}}\times \cfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}-\cfrac{1}{\sqrt{6}-\sqrt{5}}\times\cfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\cfrac{1}{\sqrt{5}-2}\times \cfrac{\sqrt{5}+2}{\sqrt{5}+2}$

$=\cfrac{3+\sqrt{8}}{(3-\sqrt8)(3+\sqrt8)}-\cfrac{\sqrt8+\sqrt7}{(\sqrt8-\sqrt7)(\sqrt8+\sqrt7)}+\cfrac{\sqrt7+\sqrt6}{(\sqrt7-\sqrt6)(\sqrt7+\sqrt6)}-\cfrac{\sqrt6+\sqrt5}{(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)}+\cfrac{\sqrt5+2}{(\sqrt5-2)(\sqrt5+2)}$$=\cfrac{3+\sqrt8}{3^2-(\sqrt8)^2}-\cfrac{\sqrt8+\sqrt7}{(\sqrt8)^2-(\sqrt7)^2}+\cfrac{\sqrt7+\sqrt6}{(\sqrt7)^2-(\sqrt6)^2}-\cfrac{\sqrt6+\sqrt5}{(\sqrt6)^2-(\sqrt5)^2}+\cfrac{\sqrt5+2}{(\sqrt5)^2-2^2}$

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

$=(3+\sqrt8)-(\sqrt8+\sqrt7)+(\sqrt7+\sqrt6)-(\sqrt6+\sqrt5)+(\sqrt5+2)$

$=3+\sqrt8-\sqrt8-\sqrt7+\sqrt7+\sqrt6-\sqrt6-\sqrt5+\sqrt5+2$

$=3+2$

$=5$

Proved.