Math, asked by surendragupta, 1 year ago

solve these questions

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Answered by dhruvrangholiyp9s5jp
1
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
4
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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