How do you rationalize #sqrt22/sqrt33#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer GiĆ³ Mar 30, 2015 Write it as: #sqrt(22)/sqrt(33)=sqrt(22/33)=# divide both by #11# #=sqrt(2/3)# Now: #sqrt(2/3)=sqrt(2)/sqrt(3)=sqrt(2)/sqrt(3)*sqrt(3)/sqrt(3)=sqrt(6)/3# Or: #sqrt(22)/sqrt(33)*sqrt(33)/sqrt(33)=sqrt(22*33)/33=sqrt((2*11)(3*11))/33=sqrt(6*11^2)/33=11/33sqrt(6)=sqrt(6)/3# 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 1265 views around the world You can reuse this answer Creative Commons License