Math, asked by shwetalohiya19p9mtqv, 1 year ago

2 root 6 minus root 5 upon 3 root 5 minus 2 root 6

Answers

Answered by DaIncredible
416
Heya !!!

Identity used :

(x + y)(x - y) =  {x}^{2}  -  {y}^{2}


 \frac{2 \sqrt{6}  -  \sqrt{5} }{3 \sqrt{5}  - 2 \sqrt{6} }  \\

On rationalizing the denominator we get,

 =  \frac{2 \sqrt{6}  -  \sqrt{5} }{3 \sqrt{5} - 2 \sqrt{6}  }  \times  \frac{3 \sqrt{5} + 2 \sqrt{6}  }{3 \sqrt{5} + 2 \sqrt{6}  }  \\  \\  =  \frac{2 \sqrt{6}(3 \sqrt{5}  + 2 \sqrt{6} ) -  \sqrt{5}(3 \sqrt{5}  + 2 \sqrt{6}   )}{ {(3 \sqrt{5} )}^{2} -  {(2 \sqrt{6}) }^{2}  }  \\  \\  =  \frac{6 \sqrt{30} + 24 - 15 - 2 \sqrt{30}  }{45 - 24}  \\  \\  =  \frac{9 + 4 \sqrt{30} }{21}

Hope this helps ☺
Answered by hukam0685
166

 \frac{2 \sqrt{6}  -  \sqrt{5} }{3 \sqrt{5} - 2 \sqrt{6}  }  \times  \frac{(3 \sqrt{5}   + 2 \sqrt{6}) }{(3 \sqrt{5} + 2 \sqrt{6})  }  \\  =  \frac{(2 \sqrt{6} -  \sqrt{5})(3 \sqrt{5}  + 2 \sqrt{6} )   }{( {3 \sqrt{5} })^{2}  - ( {2 \sqrt{6}) }^{2} }  \\  =  \frac{6 \sqrt{30} + 24 - 15 - 2 \sqrt{30}  }{45 - 24}  \\  =  \frac{4 \sqrt{30} + 9 }{21}
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