Question

simplify 1/2+√5 + 1/√5+√6 + 1/√6+√7 + 1/√7+√8 + 1/√8+√9

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Answer 1

$\huge\purple{\mid{\fbox{\tt{Answer✿࿐}}\mid}}$

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Answer 2

$\large\underline{\sf{Solution-}}$

$\sf \: \dfrac{1}{2 +\sqrt{5} }+\dfrac{1}{ \sqrt{5}+\sqrt{6} }+\dfrac{1}{ \sqrt{6} +\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{8} } + \dfrac{1}{\sqrt{8}+\sqrt{9}}$

Consider,

$\red{\bf :\longmapsto\:\dfrac{1}{2 + \sqrt{5}}}$

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{5} + 2}$

☆ On rationalizing the denominator, we get

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{5} + 2} \times \dfrac{ \sqrt{5} - 2}{ \sqrt{5} - 2}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{5} - 2}{ {( \sqrt{5} )}^{2} - {(2)}^{2} }$

$\: \: \: \: \: \: \: \: {\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{5} - 2}{5 - 4}$

$\rm \:  =  \:  \: \sqrt{5} - 2$

$\red{\bf :\longmapsto\:\dfrac{1}{2 + \sqrt{5}} = \sqrt{5} - 2 - - - - (1)}$

Consider,

$\blue{\bf :\longmapsto\:\dfrac{1}{ \sqrt{5} + \sqrt{6}}}$

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{6} + \sqrt{5} }$

☆ On rationalizing the denominator, we get

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{6} + \sqrt{5} } \times \dfrac{ \sqrt{6} - \sqrt{5} }{ \sqrt{6} - \sqrt{5}}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{6} - \sqrt{5} }{ {( \sqrt{6} )}^{2} - {( \sqrt{5} )}^{2} }$

$\: \: \: \: \: \: \: \: {\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{6} - \sqrt{5} }{6 - 5}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{6} - \sqrt{5} }{1}$

$\rm \:  =  \:  \: \sqrt{6} - \sqrt{5}$

$\blue{\bf :\longmapsto\:\dfrac{1}{ \sqrt{5} + \sqrt{6}} = \sqrt{6} - \sqrt{5} - - -(2)}$

Consider,

$\green{\bf :\longmapsto\:\dfrac{1}{ \sqrt{6} + \sqrt{7}}}$

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{7} + \sqrt{6} }$

☆ On rationalizing the denominator, we get

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{7} + \sqrt{6}} \times \dfrac{ \sqrt{7} - \sqrt{6} }{ \sqrt{7} - \sqrt{6} }$

$\rm \:  =  \:  \: \dfrac{ \sqrt{7} - \sqrt{6} }{ {( \sqrt{7} )}^{2} - {( \sqrt{6} )}^{2} }$

$\: \: \: \: \: \: \: \: {\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{7} - \sqrt{6} }{7 - 6}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{7} - \sqrt{6} }{1}$

$\rm \:  =  \:  \: \sqrt{7} - \sqrt{6}$

$\green{\bf :\longmapsto\:\dfrac{1}{ \sqrt{6} + \sqrt{7}} = \sqrt{7} - \sqrt{6} - - - (3)}$

Consider,

$\purple{\bf :\longmapsto\:\dfrac{1}{ \sqrt{7} + \sqrt{8}}}$

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{8} + \sqrt{7}}$

☆ On rationalizing the denominator, we get

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{8} + \sqrt{7}} \times \dfrac{ \sqrt{8} - \sqrt{7} }{ \sqrt{8} - \sqrt{7} }$

$\rm \:  =  \:  \: \dfrac{ \sqrt{8} - \sqrt{7} }{ {( \sqrt{8} )}^{2} - {( \sqrt{7} )}^{2} }$

$\: \: \: \: \: \: \: \: {\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{8} - \sqrt{7} }{8 - 7}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{8} - \sqrt{7} }{1}$

$\rm \:  =  \:  \: \sqrt{8} - \sqrt{7}$

$\purple{\bf :\longmapsto\:\dfrac{1}{ \sqrt{7} + \sqrt{8}} = \sqrt{8} - \sqrt{7} - - - (4)}$

Consider,

$\pink{\bf :\longmapsto\:\dfrac{1}{ \sqrt{8} + \sqrt{9}}}$

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{9} + \sqrt{8}}$

☆ On rationalizing the denominator, we get

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{9} + \sqrt{8}} \times \dfrac{ \sqrt{9} - \sqrt{8} }{ \sqrt{9} - \sqrt{8} }$

$\rm \:  =  \:  \: \dfrac{ \sqrt{9} - \sqrt{8} }{ {( \sqrt{9} )}^{2} - {( \sqrt{8} )}^{2} }$

$\: \: \: \: \: \: \: \: {\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{9} - \sqrt{8} }{1}$

$\rm \:  =  \:  \: 3 - \sqrt{8}$

$\pink{\bf :\longmapsto\:\dfrac{1}{ \sqrt{8} + \sqrt{9}} = 3 - \sqrt{8} - - - (5)}$

Hence,

$\red{\sf\:\dfrac{1}{2 +\sqrt{5} }+\dfrac{1}{ \sqrt{5}+\sqrt{6} }+\dfrac{1}{ \sqrt{6}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{8} }+\dfrac{1}{\sqrt{8}+\sqrt{9}}}$

$\rm \:=(\sqrt{5}-2)+(\sqrt{6}-\sqrt{5)}+ (\sqrt{7}-\sqrt{6})+(\sqrt{8}-\sqrt{7})+ (3-\sqrt{8})$

$\rm \:=\cancel{\sqrt{5}}-2+\cancel{\sqrt{6}}-\cancel{\sqrt{5}}+ \cancel{\sqrt{7}}-\cancel{\sqrt{6}}+\cancel{\sqrt{8}}-\cancel{\sqrt{7}}+ 3-\cancel{\sqrt{8}}$

$\rm \:  =  \:  \: 3 - 2$

$\rm \:  =  \:  \: 1$

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)