Question

$\begin{gathered} \ {if \: \frac{1}{ \sqrt{5} + 2 } + \frac{1}{2 + \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{2} } + \frac{1}{ \sqrt{2} + 1 }} \\ \\ \tt{ = a + b \sqrt{c} \: \: then \: find \: the \: value \: of \: a +b +c }\end{gathered}$

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Answer 1

Given

• $\sf {\: \dfrac{1}{ \sqrt{5} + 2 } + \dfrac{1}{2 + \sqrt{3} } + \dfrac{1}{ \sqrt{3} + \sqrt{2} } + \dfrac{1}{ \sqrt{2} + 1 } = a + b \sqrt{c} }$

To Find

• a, b, and c

Solution

$\sf {~~~ \implies \: \dfrac{1}{ \sqrt{5} + 2 } + \dfrac{1}{2 + \sqrt{3} } + \dfrac{1}{ \sqrt{3} + \sqrt{2} } + \dfrac{1}{ \sqrt{2} + 1 } = a + b \sqrt{c} }$

• Rationalize the denominator term

$\sf {~~~ \implies \: \bigg(\dfrac{1}{ \sqrt{5} + 2 } \times \dfrac{ \sqrt{5} - 2}{ \sqrt{5} - 2 } \bigg) +\bigg( \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} } \bigg)+\bigg( \dfrac{1}{ \sqrt{3} + \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }\bigg) +\bigg( \dfrac{1}{ \sqrt{2} + 1 } \times \dfrac{ \sqrt{2} - 1 }{ \sqrt{2} - 1 } \bigg)= a + b \sqrt{c} }$

• Using Identity → (a+b)(a-b) = a²-b²

$\\\sf {~~~ \implies \: \bigg(\dfrac{ \sqrt{5} - 2}{ (\sqrt{5})^{2} - {2}^{2} } \bigg) +\bigg( \dfrac{2 - \sqrt{3} }{ {2}^{2} - ( { \sqrt{3}}) ^{2} } \bigg) + \bigg( \dfrac{ \sqrt{3} - \sqrt{2} }{ (\sqrt{3})^{2} - (\sqrt{2} )^{2} } \bigg) +\bigg( \dfrac{ \sqrt{2} - 1}{ (\sqrt{2} )^{2} - {1}^{2} } \bigg)= a + b \sqrt{c} }$

$\sf {~~~ \implies \: \bigg(\dfrac{ \sqrt{5} - 2}{ 5 - 4 } \bigg) +\bigg( \dfrac{2 - \sqrt{3} }{ 4 - 3 } \bigg)+\bigg( \dfrac{ \sqrt{3} - \sqrt{2} }{ 3 - 2 } \bigg) +\bigg( \dfrac{ \sqrt{2} - 1}{ 2 - 1 } \bigg)= a + b \sqrt{c} }$

$\sf {~~~ \implies \: \sqrt{5} - 2 +2 - \sqrt{3} + \sqrt{3} - \sqrt{2} + \sqrt{2} - 1 = a + b \sqrt{c} }$

$\sf {~~~ \implies \: \sqrt{5} \: \: \cancel {- 2} \: \: \cancel {+2 } \: \: \cancel {- \sqrt{3}} \: \: \cancel { + \sqrt{3}} \: \: \cancel {- \sqrt{2}} \cancel {+ \sqrt{2}} - 1 = a + b \sqrt{c} }$

$\sf {~~~ \implies - 1 + 1 \sqrt{5} = a + b \sqrt{c} }$

• On Comparing Both side's

$\\ \sf ~~~ a = (-1) \\ \sf ~~~ b = 1 \\ \sf ~~~ c = 5$

• $\begin{gathered}\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\gray{\begin{gathered}\tiny\begin{gathered}\small{\small{\small{\small{\small{\small{\small{\small{\small{\small{\begin{gathered}\begin{gathered}\begin{gathered}\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\red{ \bigstar} \: \underline{\bf{\orange{More \: Useful \: Formula}}}\\ {\boxed{\begin{array}{cc}\dashrightarrow \sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab \\\\\dashrightarrow \sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab \\\\\dashrightarrow \sf(a + b)(a - b) = {a}^{2} - {b}^{2} \\\\\dashrightarrow \sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b) \\\\ \dashrightarrow\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b) \\ \\\dashrightarrow\sf a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab) \\\\\dashrightarrow \sf a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab )\\\\\dashrightarrow \sf{a²+b²=(a+b)²-2ab}\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}}}}}}}}}}}\end{gathered}\end{gathered}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\\ \end{gathered}$

Answer 2

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

Given that

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{5} + 2 } + \dfrac{1}{2 + \sqrt{3} } + \dfrac{1}{ \sqrt{3} + \sqrt{2} } + \dfrac{1}{ \sqrt{2} + 1 } \\ \\ \tt{ = a + b \sqrt{c}}$

Consider,

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

On rationalizing the denominator, we get

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

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

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

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

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

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

Consider,

$\red{\rm :\longmapsto\:\dfrac{1}{2 + \sqrt{3} } }$

On rationalizing the denominator, we get

$\rm \: = \: \: \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$\rm \: = \: \: \dfrac{2 - \sqrt{3} }{ {2}^{2} - {( \sqrt{3} )}^{2} }$

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

$\rm \: = \: \: \dfrac{2 - \sqrt{3} }{4 - 3}$

$\rm \: = \: \: \dfrac{2 - \sqrt{3} }{1}$

$\rm \: = \: \: 2 - \sqrt{3}$

Consider,

$\red{\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} + \sqrt{2} } }$

On rationalizing the denominator, we get

$\rm \: = \: \: \dfrac{1}{ \sqrt{3} + \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$\rm \: = \: \: \dfrac{ \sqrt{3} - \sqrt{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} }$

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

$\rm \: = \: \: \dfrac{ \sqrt{3} - \sqrt{2} }{3 - 2}$

$\rm \: = \: \: \dfrac{ \sqrt{3} - \sqrt{2} }{1}$

$\rm \: = \: \: \sqrt{3} - \sqrt{2}$

Consider,

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

On rationalizing the denominator, we get

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

$\rm \: = \: \: \dfrac{ \sqrt{2} - 1}{ {( \sqrt{2} )}^{2} - {(1)}^{2} }$

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

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

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

$\rm \: = \: \: \sqrt{2} - 1$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{5} + 2 } + \dfrac{1}{2 + \sqrt{3} } + \dfrac{1}{ \sqrt{3} + \sqrt{2} } + \dfrac{1}{ \sqrt{2} + 1 } \\ \\ \tt{ = a + b \sqrt{c}}$

On substituting all the values evaluated above, we get

$\rm :\longmapsto\:\sqrt{5} - 2 + 2 - \sqrt{3} + \sqrt{3} - \sqrt{2} + \sqrt{2} - 1 = a + b \sqrt{c}$

$\rm :\longmapsto\:\sqrt{5} - 1 = a + b \sqrt{c}$

$\rm :\longmapsto\: - 1 + \sqrt{5} = a + b \sqrt{c}$

$\rm :\longmapsto\: - 1 + 1\sqrt{5} = a + b \sqrt{c}$

On comparing, we get

$\red{\rm :\longmapsto\:a = - 1}$

$\red{\rm :\longmapsto\:b = 1}$

$\red{\rm :\longmapsto\:c = 5}$

Therefore,

$\red{\bf :\longmapsto\:a + b + c = - 1 + 1 + 5 = 5}$