Math, asked by zonunmawii, 1 year ago

1 by root3 -2 + 1 by root3 + 3

Answers

Answered by Anonymous
12
Heya ,

Solution is here ⬇⬇⬇
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given \:  \:  \frac{1}{ \sqrt{3 } - 2 } +  \frac{1}{ \sqrt{3} + 3 }   \\  \\   =  > \frac{1}{ \sqrt{3} - 2 }  =  \frac{( \sqrt{3} + 2) }{( \sqrt{3}   - 2) \sqrt{3} - 2) }  =  \frac{ \sqrt{3} + 2 }{3 - 4}  =  \\  \\  \frac{ \sqrt{3} + 2 }{ - 1}  =  - ( \sqrt{3}  + 2) +  \\  \\  =  >  \frac{1}{ \sqrt{3} + 3 }  =  \frac{( \sqrt{3} - 2) }{( \sqrt{3}  + 2)( \sqrt{3} + 2) }  =  \frac{( \sqrt{3} - 2) }{3 - 4}  =  \\  \\  \frac{( \sqrt{3} - 2) }{ - 1}  =  - ( \sqrt{3}  - 2) =  -  \sqrt{3}  + 2 \\  \\ if \:  \: we \: add \: these \:  \: two \:  \: equation \:  \:  \\  \\  =  >  \sqrt{3}  - 2 + ( -  \sqrt{3}  + 2) \\  =  >  -  \sqrt[2]{3}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: answer
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Hope it's helps you.
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Raghav3333: nice di
Anonymous: :) thanks bhai
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