Math, asked by surendragupta, 1 year ago

solve these questions

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Answered by dhruvrangholiyp9s5jp

This is perfect answer for you

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surendragupta: can't we use Identity Number 1 and 2 to solve the question

dhruvrangholiyp9s5jp: which identity

dhruvrangholiyp9s5jp: we can use it

surendragupta: if we use that identity a square + b square minus 2 a b and a square + b square + 2 a b then how could we solve it

dhruvrangholiyp9s5jp: yes it is possible to use it

surendragupta: so by using these identities what would be the answer

dhruvrangholiyp9s5jp: answer would the same that was given by me

surendragupta: thank you

Answered by Cutiepie93

Hello friends!!

Here is your answer :

Ans 1

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$x = \frac{ (\sqrt{3} + \sqrt{2})( \sqrt{3} + \sqrt{2} )}{ (\sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2}) }$

$x = \frac{ {( \sqrt{3} + \sqrt{2}) }^{2} }{(\sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2}) }$

Using identity

( a + b)² = a² + b² + 2ab

( a + b) ( a - b ) = a² - b²

$x = \frac{ {( \sqrt{3}) }^{2} + (\sqrt{2} )+ 2( \sqrt{3} )( \sqrt{2} )}{ {( \sqrt{ 3 }) }^{2} - {( \sqrt{2} )}^{2} }$

$x = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2}$

$x = 5 + 2 \sqrt{6}$

$y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$y = \frac{( \sqrt{3} - \sqrt{2})( \sqrt{3} - \sqrt{2} ) }{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2}) }$

$y = \frac{ {( \sqrt{3} - \sqrt{2}) }^{2} }{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2})}$

Using identity

( a - b)² = a² + b² - 2ab

( a + b) ( a - b ) = a² - b²

$y = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2} )}^{2} - 2( \sqrt{3})( \sqrt{2} )}{ ({ \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} }$

$y = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2}$

$y = 5 - 2 \sqrt{6}$

$= > {x}^{2} + {y}^{2}$

$= {(5 + 2 \sqrt{6}) }^{2} + {(5 - 2 \sqrt{6}) }^{2}$

Using identity

( a + b)² = a² + b² + 2ab

( a - b)² = a² + b² - 2ab

$= {(5)}^{2} + {(2 \sqrt{6} )}^{2} + 2(5)(2 \sqrt{6} ) + ( {(5)}^{2} + {(2 \sqrt{6} )}^{2} - 2(5)(2 \sqrt{6} ))$

$25 + 24 + 20\sqrt{6} + (25 + 24 - 20 \sqrt{6} )$

$25 + 24 + 20 \sqrt{6} + 25 + 24 - 20 \sqrt{6}$

$= 49 + 49$

$= 98$

Ans 2 same method like first one.

Hope it helps you.. ☺️☺️☺️☺️

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