Question

Rationalize √3 + √2 + √5

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Answer 1

Step-by-step explanation: hey square root of 3,2,5 is 3,2,5 only know! i think the answer will be 3+2+5= 10.

Answer 2

Appropriate Question :-

Rationalize

$\rm \: \dfrac{1}{ \sqrt{3} + \sqrt{2} + \sqrt{5} }$

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

Given expression is

$\rm \: \dfrac{1}{ \sqrt{3} + \sqrt{2} + \sqrt{5} }$

can be rewritten as

$\rm \: = \: \dfrac{1}{(\sqrt{3} + \sqrt{2}) + \sqrt{5} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{(\sqrt{3} + \sqrt{2}) + \sqrt{5} } \times \dfrac{( \sqrt{3} + \sqrt{2}) - \sqrt{5} }{( \sqrt{3} + \sqrt{2}) - \sqrt{5} }$

We know,

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

So, using this identity, we get

$\rm \: = \: \dfrac{ \sqrt{3} + \sqrt{2} - \sqrt{5} }{ {( \sqrt{3} + \sqrt{2} )}^{2} - {( \sqrt{5} )}^{2} }$

We know,

$\boxed{\tt{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{ \sqrt{3} + \sqrt{2} - \sqrt{5} }{ {( \sqrt{3})}^{2} + {( \sqrt{2} )}^{2} + 2. \sqrt{3}. \sqrt{2} - {( \sqrt{5}) }^{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} + \sqrt{2} - \sqrt{5} }{3 + 2 + 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} + \sqrt{2} - \sqrt{5} }{2 \sqrt{6}}$

So, on rationalizing again, we get

$\rm \: = \: \dfrac{ \sqrt{3} + \sqrt{2} - \sqrt{5} }{2 \sqrt{6}} \times \dfrac{ \sqrt{6} }{ \sqrt{6} }$

$\rm \: = \: \dfrac{ \sqrt{18} + \sqrt{12} - \sqrt{30} }{2 {( \sqrt{6})}^{2} }$

$\rm \: = \: \dfrac{ \sqrt{2 \times 3 \times 3} + \sqrt{2 \times 2 \times 3} - \sqrt{30} }{2 \times 6 }$

$\rm \: = \: \dfrac{3 \sqrt{2} + 2 \sqrt{3} - \sqrt{30} }{12}$

Hence,

$\boxed{\tt{ \: \: \: \rm \:\dfrac{1}{ \sqrt{3} + \sqrt{2} + \sqrt{5} } = \: \dfrac{3 \sqrt{2} + 2 \sqrt{3} - \sqrt{30} }{12} \: \: \: \: }}$

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ADDITIONAL INFORMATION

MORE ALGEBRAIC IDENTITIES

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

$\boxed{\tt{ \: \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \: }}$

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

$\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: }}$

$\boxed{\tt{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x + y) \: }}$

$\boxed{\tt{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2}) \: }}$

$\boxed{\tt{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}) \: }}$

$\boxed{\tt{ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \: }}$

$\boxed{\tt{ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x - y) \: }}$