Question

if 1/√5+2+1/2+√3+1/√3+√2+1/√2+1 = a+b√c, then find the value of a+b+c.

please give the full explanation of the answer. it's a humble request to all
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Answer 1

I solved it till here. hope it helps you.

i think squaring both sides will bring some thing new.

It will give you the value of a square + BC you can transfer BC on another site and it will give you the value of a square. you can transfer bc on another side and then transfer the square root on another side and get the value of a and then put the value of a in the given equation and get the value of b and c respectively.

and finally add them.

Answer 2

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

Given that

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

Consider,

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

On rationalizing the denominator, we get

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

We know,

$\underbrace{\boxed{ \tt{(x + y)(x - y) = {x}^{2} - {y}^{2} }}}$

So, using this identity, we get

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

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

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

$\rm \:  =  \:  \: \sqrt{5} - 2$

Consider,

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

On rationalizing the denominator, we get

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

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

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

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

$\rm \:  =  \:  \: 2 - \sqrt{3}$

Consider,

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

On rationalizing the denominator, we get

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

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

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

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

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

Consider,

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

On rationalizing the denominator, we get

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

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

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

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

$\rm \:  =  \:  \: \sqrt{2} - 1$

Now, Consider the given expression,

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

can be rewritten as by substituting the values evaluated above,

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

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

can be re-arranged as

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

So, on comparing we get

$\bf\implies \:a = - 1$

$\bf\implies \:b = 1$

$\bf\implies \:c = 5$