How do you simplify # (1-sqrt2)/(2+sqrt6)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Mark D. Apr 10, 2018 #=(2-3sqrt(6)+2sqrt(3))/(-2)# Explanation: Rationalise : #(1-sqrt(2))/(2+sqrt(6))##xx##(2-sqrt(6))/(2-sqrt(6))# #=># #=(2-sqrt(6)-2sqrt(6)+sqrt(12))/(4-2sqrt(6)+2sqrt(6)-6# #=>#. #=(2-3sqrt(6)+2sqrt(3))/(-2)# 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 1019 views around the world You can reuse this answer Creative Commons License