How do you simplify #3/(6-sqrt5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer GiĆ³ Jun 8, 2015 You can multiply and divide by #6+sqrt(5)# to get: #3/(6-sqrt(5))*(6+sqrt(5))/(6+sqrt(5))=(3(6+sqrt(5)))/(36-5)=(3(6+sqrt(5)))/31# Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 969 views around the world You can reuse this answer Creative Commons License