Math, asked by sree999, 1 year ago

pls answer the question as soon as possible!!

Attachments:

sree999: fyn,who are u
Aditya0000: I am new on sree
Aditya0000: I can be your friend
sree999: oo
sree999: ya
Aditya0000: yaaa
siddhartharao77: Please dont chat here...
sree999: sorry
Anonymous: ya.... don't chat here plz
sree999: ok fyn no more comments will be there

Answers

Answered by Anonymous
7
Hii..

Answer is in attachment.

hope this helps !
Attachments:

sree999: hmm..
sree999: why
Anonymous: :)
sree999: i am girl
Anonymous: are u boy or girl ??!!
sree999: girl
sree999: why u have a doubt
Anonymous: no ... sorry i told u bro without seeing ur pic :(
sree999: kk
sree999: no prblm
Answered by siddhartharao77
6
= \ \textgreater \   \frac{5 +  \sqrt{3} }{7 + 4 \sqrt{3} }

= \ \textgreater \   \frac{(5 +  \sqrt{3})(7 - 4 \sqrt{3} ) }{(7 + 4 \sqrt{3})(7 - 4 \sqrt{3})  }

= \ \textgreater \   \frac{ 5*7 - 5 * 4 \sqrt{3} +  \sqrt{3} * 7 -  \sqrt{3} * 4 \sqrt{3}}{(7)^2 - (4 \sqrt{3})^2 }

= \ \textgreater \   \frac{35 - 20 \sqrt{3} + 7 \sqrt{3} - 12  }{49 - 48}

= \ \textgreater \   \frac{35 - 13 \sqrt{3} - 12 }{1}

= \ \textgreater \  23 - 13 \sqrt{3}


Hope this helps!

sree999: thank you
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