How do you rationalize the denominator and simplify #(2+sqrt3)/sqrt3#?

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Answer 1 · 2018-03-13T12:15:22

See a solution process below:

Multiply by #sqrt(3)/sqrt(3)# giving:

#sqrt(3)/sqrt(3) xx (2 + sqrt(3))/sqrt(3) =>#

#(sqrt(3)(2 + sqrt(3)))/(sqrt(3) xx sqrt(3)) =>#

#(2sqrt(3) + sqrt(3)sqrt(3))/3 =>#

#(2sqrt(3) + 3)/3#

Or

#(2sqrt(3))/3 + 3/3 =>#

#2/3sqrt(3) + 1#