How do you simplify #sqrt3 div(sqrt12 - sqrt5)#?

1 Answer
Bhavin B.
May 15, 2018

#(sqrt3)/(2sqrt3-sqrt5)#

Explanation:

First, we find the factors of #sqrt12# and simplify it further.

#sqrt12# = #sqrt(4 xx 3# = #sqrt4xxsqrt3# = #2xxsqrt3#

#(sqrt3)/(sqrt12 - sqrt5)# ----->we know that #sqrt12# is #2xxsqrt3# and substitute that for #sqrt12# as shown below:

#(sqrt3)/(2xxsqrt3 -sqrt5)# = #(sqrt3)/(2sqrt3-sqrt5)#

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