How do you rationalize the denominator and simplify #sqrt(5/3)#?

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Answer 1 · 2018-06-18T17:46:59

#sqrt15/3#

#sqrt(5/3)#

use the rule: #sqrt(a/b) = sqrta/sqrtb#

#sqrt5/sqrt3#

now multiply by 1 in the form #a/a# to rationalize:

#sqrt5/sqrt3 * sqrt3/sqrt3 = sqrt15/sqrt9=sqrt15/3#

Answer 2 · 2018-06-18T17:49:04

#sqrt15/3#

#sqrt(5/3) rArr sqrt5/sqrt3#

#sqrt5/sqrt3 xx sqrt3/sqrt3#

#rArr =sqrt15/sqrt9#

#rArr sqrt15/3#